Potpuno riješeni zadatci za vježbu PDF-varijanta - MIM-Sraga

M-1-Kub zbroja i kub razlike
Rješenja + kompletni postupak
40.
Koristeći se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2 3
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+ 3AB −B
izračunaj:
3 3
( x+ y) ( x−
y) ( y−x)
1) 2) 3)
( x−
) ( −x) ( x+
)
3 3 3
4) 1 5) 1 6) 1
3 3
( x−
) ( −x) ( x+
)
7) 2 8) 2 9) 3
3
( a+ b)
11) ( 2a−3b) 12) ( 3x−5y)
10) 2
3 3
3 3
( x+ y) ( ab−
c) ( ab+
c)
13) 5 2 14) 5 15) 3 5
( x + ) ( x − y ) ( x − y )
16) 1 17) 18)
2 3 2 3 3 3 3 3
3 3
⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞
19) ⎜x− ⎟ 20) ⎜2x+
⎟ 21) ⎜3x−
⎟
⎝ 3⎠ ⎝ 2⎠ ⎝ 3⎠
3 3
⎛1 2⎞ ⎛1 1 ⎞ ⎛ 2 3 1⎞
22) ⎜ x+ ⎟ 23) ⎜ ab+
c⎟ 24) ⎜a b + ⎟
⎝2 3⎠ ⎝3 4 ⎠ ⎝ 3⎠
3
⎛ 2 3⎞ ⎛
25) 26) 0.2
3 4 5 ⎞
⎜ x − y ⎟ ⎜ x+
y⎟
⎝4 3 ⎠ ⎝ 6 ⎠
( x + )
3 3 2 3
( x+ ) ( x−
y) ( x+
y )
2 3 3 3 4 3 4 2 3 3
( xy + z ) ( xy −z ) ( x y −z
)
4 2 3 3 2 3 2 2 3 2
( ab+ ab) ( xy+
) ( xy−
xy)
2 3 2 3 2 3 5 3 2 3 4 3
( a + a b ) ( a − b c ) ( x + y z )
−1 3 −2 3 −3 2 3
( x + x) ( x − x) ( x + x )
m n 3 n n 3
n n 3
( − ) ( − )
45) ( 2 + 3 )
n n 3 n n+ 1 3
n+
1 n−1 3
( a + b ) ( a −a ) ( a + a )
3
27) 0.1 10
28) 2 0.1 29) 0.1 0.2 30) 5 0.4
31) 32) 33)
34) 35) 2 3 36) 3
37) 2 38) 3 2 39) 5
40) 41) 42)
43) 2 2 44) 3 2
46) 47) 48)
( a− ) + ( a+ ) ( a− ) − ( a+
)
3 3 3 3
49) 1 1 50) 1 1
( x+ y) −( x− y) ( x+ y) + ( x−
y)
3 3 3 3
51) 52)
( x+ ) − ( x− ) ( x+ ) − ( −x)
3 3 3 3
53) 2 3 3 1 54) 3 1 5 2
( x+ y) − ( x+ y) ( x− ) − x( x+
)
3 3 3 2
55) 3 2 4 2 56) 1 2
2 3 3 3 3 3
( − ) − ( + ) ( + ) − ( − ) − ( + )
57) x x 2 x 2 58) 2 x y 5 x y x y
3
3
3
3
3
3
2
1
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
( )
3 3 2 2 3 3 3 2
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+
3AB
3 3 2 2 3 3 2 2 3
1) x+ y = x + 3⋅x ⋅ y+ 3⋅x⋅ y + y = x + 3x y+ 3xy + y
−B
2 3
uputa:
( )
3 3 2 2 3
x+ y = x + 3⋅x ⋅ y+ 3⋅x⋅ y + y =
↓ ↓ ↑ ↑ ↑ ↑
( )
3 3 2 2 3
A+ B = A + 3⋅A ⋅ B+ 3⋅AB ⋅ + B
x + 3x y+ 3xy + y
3 2 2 3
( )
( )
3 3 2 2 3 3 2 2 3
2) x− y = x −3⋅x ⋅ y+ 3⋅x⋅y − y = x − 3x y+ 3xy − y
↓ ↓ ↑ ↑ ↑ ↑
3 3 2 2 3
A− B = A −3⋅A ⋅ B+ 3⋅AB ⋅ − B
( )
3 3 2 2 3 3 2 2 3
3) y− x = y −3⋅y ⋅ x+ 3⋅y⋅x − x = y − 3y x+ 3yx −x
( )
3 3 2 2 3 3 2
4) x− 1 = x −3⋅x ⋅ 1+ 3⋅x⋅1 − 1 = x − 3x + 3x−1
( )
5) 1− x = 1 − 3⋅1 ⋅ x + 3⋅1⋅x − x = 1− 3x+ 3x −x
( )
3 3 2 2 3 2 3
↓ ↓ ↑ ↑ ↑ ↑
3 3 2 2 3
A− B = A −3⋅A ⋅ B+ 3⋅AB ⋅ − B
( )
3 3 2 2 3 3 2
6) x+ 1 = x + 3⋅x ⋅ 1+ 3⋅x⋅ 1 + 1 = x + 3x + 3x+
1
( )
( )
3 3 2 2 3 3 2
3 2
7) x− 2 = x − 3⋅x ⋅ 2 + 3⋅x⋅2 − 2 = x − 6x + 3⋅4⋅x− 8 = x − 6x
+ 12x−8
↓ ↓ ↑ ↑ ↑ ↑
3 3 2 2 3
A− B = A −3⋅A ⋅ B+ 3⋅AB ⋅ − B
( x)
3 3
8) 2 − = 2 −3⋅2
⋅ x+ 32 ⋅ ⋅x − x = 8−34 ⋅ ⋅ x+ 6x − x = 8−12x+ 6x −x
2 2 3 2 3
2
2 3
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) 3 3
( )
( )
3 3 2 2 3 3 3 2
A+ B = A + A B+ AB + B A− B = A −3A B+ 3AB −B
3 3 2 2 3
9) x+ 3 = x + 3⋅x ⋅ 3 + 3⋅x⋅ 3 + 3 =
9 3 9 27
3 2
= x + x + ⋅ ⋅ x+ =
3 2
= x + x + x+
9 27 27
2 3
3 3 2
( a+ b)
= a + ⋅ a 2b 3 a ( 2b) ( 2b)
10) 2 3
2 3
⋅ + ⋅ ⋅ + =
3 2 2 2 3 3
= a + 6a b+ 3a⋅2 ⋅ b + 2 ⋅ b =
= a + 6a b+ 12ab + 8b
3 2 2 3
( a b) ( a) ( a) b a ( b) ( b)
3 3 2 2 3
11) 2 − 3 = 2 −3⋅ 2 ⋅ 3 + 3⋅2 ⋅ 3 − 3 =
3 3 2 2 2 2 3 3
= ⋅ − ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅ =
2 a 3 2 a 3 b 6 a 3 b 3 b
8a 343a b 69a b 27b
3 2 2 3
= − ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − =
= 8a − 36a b+ 54ab −27b
3 2 2 3
( )
3 3 2
A − B = A − 3⋅A ⋅ B + 3⋅A⋅
B
2 3
↑ ↑ ↓ ↓ ↓ ↓
( x y) ( x) ( x) y x ( y) ( y)
3 3 2 2 3
12) 3 − 5 = 3 −3⋅ 3 ⋅ 5 + 3⋅3 ⋅ 5 − 5 =
3 3 2 2 2 2 3 3
= ⋅ − ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅ =
3 x 3 3 x 5 y 9 x 5 y 5 y
3 2 2 3
= − ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − =
27x 3 9 5 x y 9 25 x y 125y
= 27x − 135x y+ 225xy −125y
3 2 2 3
−
B
( )
3 3 2 2 3
A + B = A + 3⋅A ⋅ B + 3⋅A⋅ B + B
↑ ↑ ↓ ↓ ↓ ↓
3 3 2
( x+ y) = ( x) + ⋅( x) ⋅ y+ ⋅ x⋅ ( y)
( 2y)
13) 5 2 5 3 5 2 3 5 2
2 3
3 3 2 2 2 2 3 3
= 5 ⋅ x + 3⋅5 ⋅x ⋅2⋅ y+ 15⋅x⋅2 ⋅ y + 2 ⋅ y =
3 2 2 3
= 125x + 3⋅25⋅2⋅x ⋅ y+ 15⋅4⋅x⋅ y + 8y
=
= 125x + 150x y+ 60xy + 8y
3 2 2 3
+ =
3
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+
3AB
( ab c) ( ab) ( ab) c ab ( c) ( c)
3 3 2 2 3
14) − 5 = −3⋅ ⋅ 5 + 3⋅ ⋅ 5 − 5 =
ab 3 a b 5 c 3 ab5 c 5 c
3 3 2 2 2 2 2 3 3
= − ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ − ⋅ =
3 3 2 2 2 2 3
= ab −35 ⋅ ⋅a⋅b⋅ c + 325 ⋅ ⋅abc ⋅ ⋅ − 125c
=
= a b − 15a b c + 75abc −125c
3 3 2 2 2 2 3
−B
2 3
( ab c) ( ab) ( ab) c ab ( c) ( c)
3 3 2 2 3
15) 3 + 5 = 3 −3⋅ 3 ⋅ 5 + 3⋅3 ⋅ 5 + 5 =
3 a b 3 9 a b 5 c 9 a b 25 c 5 c
3 3 3 2 2 2 3 3
= ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ + ⋅ =
27a b 3 9 5 a b c 9 25 abc 125c
3 3 2 2 2 3
= − ⋅ ⋅ ⋅ + ⋅ ⋅ + =
3 3 2 2
27ab
135abc
2
= − +
25abc
+ 125c
2 3
( ) ( ) ( )
2 3 2 3 2 2 2 2 3
16) x + 1 = x + 3⋅ x ⋅ 1+ 3⋅x
⋅ 1 + 1 =
23 ⋅ 22 ⋅ 2
= + 3⋅ + 3⋅ ⋅ 1+ 1=
x x x
6 4 2
= + + +
x 3x 3x
1
( x y ) ( x ) ( x ) y x ( y ) ( y )
2 3 3 2 3 2 2 3 2 3 2 3 3
17) − = −3⋅ ⋅ + 3⋅ ⋅ − =
23 ⋅ 22 ⋅ 3 2 32 ⋅ 33 ⋅
= −3⋅ ⋅ + 3⋅ ⋅ − =
x x y x y y
= x − 3x y + 3x y − y
6 4 3 2 6 9
( x y ) ( x ) ( x ) y x ( y ) ( y )
3 3 3 3 3 3 2 3 3 3 2 3 3
18) − = −3⋅ ⋅ + 3⋅ ⋅ + =
33 ⋅ 32 ⋅ 3 3 32 ⋅ 33 ⋅
= x −3⋅x ⋅ y + 3⋅x ⋅ y + y =
= x − 3x y + 3x y + y
9 6 3 3 6 9
3 2 3
3 2
⎛ 1⎞ 1 ⎛1⎞ ⎛1⎞
19) ⎜x− ⎟ = x −3⋅x ⋅ + 3⋅x⋅⎜ ⎟ − ⎜ ⎟ =
⎝ 3⎠ 3 ⎝3⎠ ⎝3⎠
3
3 1 2 1 1 1 3 3⋅1 1
= x −3⋅ ⋅ x + 3⋅ ⋅x− = → 3⋅ = = =
3
3 9 3
9 9 3⋅3 3
3 2 1 1
= x − x + x−
3 27
4
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2 3
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+ 3AB −B
3 2 3
3 2
⎛ 1⎞ 1 ⎛1⎞ ⎛1⎞
20) ⎜2x+ ⎟ = ( 2x) + 3⋅( 2x)
⋅ + 3⋅2x⋅ ⎜ ⎟ + ⎜ ⎟ =
⎝ 2⎠ 2 ⎝2⎠ ⎝2⎠
2 3
3 3 2 2 1 1 1
= 2 ⋅ x + 3⋅2 ⋅x
⋅ + 6⋅x⋅ + =
2 3
2 2 2
3 1 2 3 1 1
1 2
= 8x
+⋅3⋅4⋅ ⋅ x + 6⋅
⋅ x + = →34 ⋅ ⋅ = 3⋅
4⋅
2 4 8
2
3 2 3 1
= 8x + 6x + x+
2 8
2
1
2
1
321 ⋅ ⋅
= = 6
1
3 2 3
3 2
⎛ 1⎞ 1 ⎛1⎞ ⎛1⎞
21) ⎜3x− ⎟ = ( 3x) −3⋅( 3x)
⋅ + 3⋅3x⋅⎜ ⎟ − ⎜ ⎟ =
⎝ 3⎠ 3 ⎝3⎠ ⎝3⎠
2 3
3 3 2 2 1 1 1
= 3 ⋅x −3⋅3 ⋅x ⋅ + 9⋅x⋅ − =
2 3
3 3 3
3 1 2 1 1
1 1 119 ⋅⋅
= 27x −3⋅ ⋅9⋅ x + 9⋅ ⋅x − = → 3⋅ ⋅ 9 = 3⋅
⋅ 9 = = 9
3 9 27
3 3 1
3 2 1
= 27x − 9x + x−
27
( )
3 3 2 2 3
A + B = A + 3⋅A ⋅ B + 3⋅A⋅ B + B
↑ ↑ ↓ ↓ ↓ ↓
3 3 2 2 3
⎛1 2⎞ ⎛1 ⎞ ⎛1 ⎞ 2 1 ⎛2⎞ ⎛2⎞
22) ⎜ x+ ⎟ = ⎜ x⎟ + 3⋅⎜ x⎟ ⋅ + 3⋅ x⋅ ⎜ ⎟ + ⎜ ⎟ =
⎝2 3⎠ ⎝2 ⎠ ⎝2 ⎠ 3 2 ⎝3⎠ ⎝3⎠
3 2 2 3
1 3 1 2 2 3 2 2
= ⋅ x + 3⋅ ⋅x ⋅ + ⋅x⋅ + =
3 2 2 3
2 2 3 2 3 3
1 3 1 2 2 3 4 8
= x + 3⋅ ⋅ ⋅ x + ⋅ ⋅ x + =
8 4 3 2 9 27
1
x
8
3
= +
3 ⋅
1 2
⋅
4 3
3 2
= x + x + x+
2
1
1
2 3
⋅ x +
1 1 2 8
8 2 3 27
4
2 ⋅
1
2
9 3
8
⋅ x + =
27
5
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2 3
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+ 3AB −B
3 3 2 2
⎛1 1 ⎞ ⎛1 ⎞ ⎛1 ⎞ 1 1 ⎛1 ⎞ ⎛1
⎞
23) ⎜ ab+ c⎟ = ⎜ ab⎟ + 3⋅⎜ ab⎟ ⋅ c + 3⋅ ab⋅ ⎜ c⎟ + ⎜ c⎟
=
⎝3 4 ⎠ ⎝3 ⎠ ⎝3 ⎠ 4 3 ⎝4 ⎠ ⎝4
⎠
3 2 2 3
1 3 3 1 2 2 1 3 1 2 1 3
= ⋅a ⋅ b + 3⋅ ⋅a ⋅b ⋅ ⋅ c + ⋅a⋅b⋅ ⋅ c + ⋅ c =
3 2 2 3
3 3 4 3 4 4
1 3 3 1 1 2 2 1 2 1 3 1 1
= a b + 3⋅ ⋅ ⋅a ⋅b ⋅ c + abc + c = →3⋅ ⋅ =
27 9 4 16 64
9 4
1 1 1 1
= ab + abc+ abc + c
27 12 16 64
3 3 2 2 2 3
3
3 ⋅11
⋅
3 ⋅34 ⋅
=
1
12
3 2 3
2 3 1 2 3 3 2 3 2 1 2 3 1 1
( ) ( )
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
24) ⎜ab + ⎟ = ab + 3⋅ ab ⋅ + 3⋅ab⋅ ⎜ ⎟ + ⎜ ⎟ =
⎝ 3⎠ 3 ⎝3⎠ ⎝3⎠
1 1 1
( ) ( ) ( ) ( )
3 3 3
1 1
a b a b 3 a b 9 27
2 3
2 3 3 3 2 2 3 2 2 3
= a ⋅ b + 3⋅ ⋅ a ⋅ b + 3⋅a ⋅b
⋅ + =
2 3
23 ⋅ 33 ⋅ 22 ⋅ 32 ⋅
2 3
= ⋅ + ⋅ + ⋅ ⋅ ⋅ + =
1 1
ab ab ab
3 27
6 9 4 6 2 3
= + + +
3 3 2 2 3
⎛3 2 4 3⎞ ⎛3 2⎞ ⎛3 2⎞ 4 3 3 2 ⎛4 3⎞ ⎛4
3⎞
25) ⎜ x − y ⎟ = ⎜ x ⎟ −3⋅⎜ x ⎟ ⋅ y + 3⋅ x ⋅⎜ y ⎟ − ⎜ y ⎟ =
⎝4 3 ⎠ ⎝4 ⎠ ⎝4 ⎠ 3 4 ⎝3 ⎠ ⎝3
⎠
3 2 2 3
3 2 3 3 2 2 4 3 9 2 4 3 2 4 3 3
= ⋅
3 ( x ) −3 ⋅ ⋅
2 ( x ) . ⋅ y + ⋅x ⋅ ⋅
2 ( y ) − ⋅
3 ( y ) =
4 4 3 4 3 3
27 23 ⋅ 9 4 22 ⋅ 3 9 16 2 32 ⋅ 64 33 ⋅
= ⋅x −3⋅ ⋅ ⋅x ⋅ y + ⋅ ⋅x ⋅y
− y =
64 16 3 4 9 27
27 3 ⋅9⋅4
⋅x
⋅ y +
64 4⋅ 4⋅
3
6
= x −
4 3 9
27 9 64
= x − x y + 4x y − y
64 4 27
6 4 3 2 6 9
⋅ 4 ⋅4
2 6 64 9
⋅x
⋅y
− y =
4 ⋅ 9 27
6
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+
3AB
−B
3
3 3
⎛ 5 ⎞ ⎛ 2 5 ⎞
26) ⎜0.2x+ y⎟ = ⎜ x+ y⎟
= → decimalni broj ( 0.2 ) pretvorimo u razlomak
⎝ 6 ⎠ ⎝10 6 ⎠
⎛1 5 ⎞
= ⎜ x+ y⎟
=
⎝5 6 ⎠
3
3 2 2 3
⎛1 ⎞ ⎛1 ⎞ 5 1 ⎛5 ⎞ ⎛5
⎞
= ⎜ x⎟ + 3⋅⎜ x⎟ ⋅ y+ 3⋅ x⋅ ⎜ y⎟ + ⎜ y⎟
=
⎝5 ⎠ ⎝5 ⎠ 6 5 ⎝6 ⎠ ⎝6
⎠
3 2 2 3
1 3 1 2 5 3 5 2 5 3
= ⋅ x + 3⋅ ⋅x ⋅ ⋅ y + ⋅x⋅ ⋅ y + ⋅ y =
3 2 2 3
5 5 6 5 6 6
1 1 5 3 25 125
x 3 x y x y y
125 25 6 5 36 216
3 2 2 3
= + ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + =
1 3⋅⋅
15
x
125 5⋅ 5⋅
3 ⋅2 ⋅ xy+
3
= +
2 3
1 1 5 125
= x + x y+ xy + y
125 10 12 216
3 2 2 3
⋅ 5 ⋅5
2 125 3
⋅ xy + y =
5 ⋅ 3 ⋅12
216
( ) ( ) ( )
3 3 2 2 3
27) 0.1x+ 10 = 0.1x + 3⋅ 0.1x ⋅ 10 + 3⋅0.1x⋅ 10 + 10 =
= ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ + =
3 3 2 2
0.1 x 3 0.1 x 10 0.3 x 100 1000
= + ⋅ ⋅ ⋅ + ⋅ ⋅ + =
3 2
0.001x 3 0.01 10 x 0.3 100 x 1000
= + + +
3 2
0.001x 0.3x 30x
1000
( )
II način da decimalni broj 0.1 pretvorimo u razlomak
3 ⎛ 1 ⎞
0.1 + 10 = ⎜ x+ 10⎟
=
⎝10
⎠
( x )
3
3 2
⎛ 1 ⎞ ⎛ 1 ⎞ 1 2 3
= x + ⋅ x ⋅ + ⋅ x⋅ + =
⎜ ⎟ 3 ⎜ ⎟ 10 3 10 10
⎝10 ⎠ ⎝10 ⎠ 10
3
1 3 1 2 1
3
= ⋅ x + 3⋅ ⋅x ⋅ 10 + 3⋅ ⋅100⋅ x+ 1000 =
10 100 10
1 3 3⋅⋅ 1 10 2 3⋅10⋅10
= x + ⋅ x + ⋅ x + 1000 =
1000 10⋅10 10
1 3 2
x 30 1000
1000 10 x x
3
= +
+ + 7
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40. Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2 3
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+ 3AB −B
3 3 2 2 3
3
( x+ ) = ( x) + ⋅( x) ⋅ + ⋅ x⋅ + =
28) 2 0.1 2 3 2 0.1 3 2 0.1 0.1
= ⋅ x + ⋅
3 3 2 2
2 3 2
⋅x
⋅ 0.1 + 6⋅x⋅ 0.01 + 0.001 =
3 2
= 8x + 3⋅4⋅0.1⋅ x + 6⋅0.01⋅ x + 0.001 =
= x + x + x +
3 2
8 1.2 0.06 0.001
→ 0.1 = 0.1⋅0.
1⋅ 0.1 = 0.001
( x y) ( x) ( x) y x ( y) ( y)
3 3 2 2 3
29) 0.1 − 0.2 = 0.1 − 3⋅ 0.1 ⋅ 0.2 + 3⋅0.1 ⋅ 0.2 − 0.2 =
0.1 x 3 0.1 x 0.2 y 0.3 x 0.2 y 0.2 y
3 3 2 2 2 2 3 3
= ⋅ − ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅ =
3 2
x x y
= 0.001 − 3⋅0.01⋅0.2⋅ ⋅
0.3 0.04 x y 0.008 y
2 3
+ ⋅ ⋅ ⋅ − =
= 0.001x − 0.006x y + 0.012xy − 0.008 y
3 2 2 3
3
2 3 ⎛
2⎞
30) 5x 0.4y 5x y
0.4
4 4 2
+ = ⎜ + ⎟ = → pretvorimo u: =
⎝ 10 ⎠
10 5
( )
3
⎛ 2 2 ⎞
= 5x+ y =
⎜
⎝
5
⎟
⎠
2 3
3 2 2 2 ⎛2 2⎞ ⎛2
2⎞
( ) ( )
= 5x + 3⋅ 5x ⋅ y + 3⋅5x⋅ ⎜ y ⎟ + ⎜ y ⎟ =
5 ⎝5 ⎠ ⎝5
⎠
2
3
3 3 2 2 2 2 2 2 2 2 2 3
= 5 ⋅ x + 3⋅5 ⋅x ⋅ ⋅ y + 15⋅x⋅
⋅ ( y
2 ) + ⋅
3 ( y ) =
5 5 5
3 2 2 2 4 2⋅2 8 2⋅3
= 125x + 3⋅25⋅ ⋅x ⋅ y + 15⋅ ⋅x⋅ y + ⋅ y =
5 25 125
125x
3
= +
35 ⋅ ⋅5⋅2
2 2 35 ⋅ ⋅4
4 8 6
⋅ xy + ⋅ xy + y =
5
5 ⋅5
125
12 8
= 125x + 30x y + xy + y
5 125
3 2 2 4 6
( xy z ) ( xy ) ( xy ) z xy ( z ) ( z )
⋅ ⋅
x ( y ) 3 x ( y ) z 3xy z z
2 3 3 2 3 2 2 3 2 3 2 3 3
31) + = + 3⋅ ⋅ + 3⋅ ⋅ + =
3 2 3 2 2 2 3 2 3 2 3 3
= ⋅ + ⋅ ⋅ ⋅ + ⋅ + =
x y 3x y z 3xy z z
3 2⋅3 2 2⋅2 3 2 6 9
= ⋅ + ⋅ ⋅ + + =
= xy + 3xyz + 3xyz + z
3 6 2 4 3 2 6 9
8
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2 3
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+ 3AB −B
( xy z ) ( xy ) ( xy ) z xy ( z ) ( z )
3 3 3 2 3 2 4 3 4⋅2 4
= x ⋅( y ) −3⋅x ⋅( y ) ⋅ z + 3xy ⋅z −z
3 4 3 3 3 3 2 4 3 4 2 4 3
32) − = −3⋅ ⋅ + 3⋅ ⋅ − =
xy 3x y z 3xyz z
3 3⋅3 2 3⋅2 4 3 8 12
= − ⋅ ⋅ + − =
= xy − 3xyz + 3xyz −z
3 9 2 6 4 3 8 12
⋅3
=
( xy z) ( xy) ( xy) z xy ( z) ( z)
( ) ( ) 3 ( ) ( ) 3
4 2 3 3 4 2 3 4 2 2 3 4 2 3 2 3 3
33) − = −3⋅ ⋅ + 3⋅ ⋅ − =
4 3 2 3 4 2 2 2 3 4 2 3⋅2 3⋅3
= ⋅ − ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − =
x y x y z x y z z
x y 3 x y z 3x y z z
43 ⋅ 23 ⋅ 42 ⋅ 22 ⋅ 3 4 2 6 9
= ⋅ − ⋅ ⋅ ⋅ + − =
= x y − 3x y z + 3x y z
12 6 8 4 3 4 2 6
− z
9
4 2 3 3 4 3 4 2 2 3 4 2 3 2 2 3 3
( ab + a b ) = ( ab ) + ⋅( ab ) ⋅ a b + ⋅ab ⋅ ( a b ) + ( a b ) =
= a ⋅ ( b ) + 3⋅a ⋅( b ) ⋅a⋅2⋅ b + 3⋅a⋅b ⋅( a ) ⋅ ( b ) + ( a ) ⋅( b )
34) 3 3
3 4 3 2 4 2 3 4 2 2 3 2 2 3 3 3
3 4⋅3 2 1 4⋅2 3 1 2⋅2 4 3⋅2 2⋅3 3⋅3
= a ⋅ b + ⋅a ⋅a ⋅b ⋅ b + ⋅a ⋅a ⋅b ⋅ b + a ⋅ b =
3 12 2+ 1 8+ 3 1+ 4 4+
6 6 9
= a b + 3⋅a ⋅ b + 3⋅a ⋅ b + a b =
= ab + 3ab
3 12 3 11
3 3
+ 3ab
+ ab
5 10 6 9
=
( ) ( ) ( )
( ) ( ) ( ) ( )
2 3 2 2 3 3 2 3 2 2 3 2 3
35) 2xy+ 3 = 2xy + 3⋅ 2xy ⋅ 3+ 3⋅2xy⋅ 3 + 3 =
3 2 3 3 3 2 2 2 3 2 2 3
= 2 ⋅ ⋅ + 3⋅2 ⋅ ⋅ ⋅ 3+ 6⋅9⋅ + 27 =
x y x y x y
23 ⋅ 33 ⋅ 22 ⋅ 32 ⋅ 2 3
= 8⋅ ⋅ + 3⋅4⋅3⋅ ⋅ + 54 + 27 =
x y x y x y
= xy+ xy + xy+
6 9 4 6 2 3
8 36 54 27
2 3 2 2 2 3 3 2 3 2 2 2 3 2 2 2 3
( xy− xy) = ( xy) − ⋅( xy)
⋅ xy + ⋅xy⋅( xy) − ( 3xy
) =
( x ) ( y ) ( x ) ( y ) x y x y x ( y )
36) 3 3 3 3 3
= ⋅ −33 ⋅ ⋅ ⋅ ⋅ ⋅ + 3⋅ ⋅ ⋅3⋅ ⋅ −3
= x ⋅y −9⋅x ⋅x ⋅y ⋅ y + 3⋅9⋅x ⋅x ⋅y ⋅y −27⋅x ⋅y
2⋅3 3⋅3 2⋅2 1 3⋅2 2 2 2 3 2⋅2 3 2⋅3
6 9 4+ 1 6+ 2 2+ 2 3+
4 3 6
= − ⋅ ⋅ + ⋅ ⋅ − =
xy 9 x y 27 x y 27xy
= xy − 9xy+ 27xy −27xy
6 9 5 8 4 7 3 6
( y )
2 3 3 3 2 2 3 2 2 2 3 2 2 2 2 3
⋅ 3 ⋅ 2 3 =
=
x
9
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2 3
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+ 3AB −B
2 3 2 3 2 3 2 2 3 2 2 3 2 2 3 2 3
( a + ab ) = ( a ) + ⋅( a ) ⋅ ab + ⋅a ⋅ ( ab ) + ( ab ) =
23 22 3 2 2
a ⋅
⋅
= + 32 ⋅ ⋅a ⋅a ⋅ b + 3⋅ a ⋅2 ⋅( a ) ⋅ ( b ) + 2 ⋅( a ) ⋅( b )
37) 2 3 2 3 2 2
2 3 2 2 2 3 3 3 2 3
6 4+ 3 2 2 3⋅2 2⋅2 3⋅3 2⋅3
= + 6⋅ ⋅ + 3⋅4⋅ ⋅ ⋅ + 8⋅ ⋅ =
a a b a a b a b
6 7 2 2+
6 4 9 6
= + 6 + 12⋅ ⋅ + 8 =
a a b a b a b
= a + 6a b + 12a b + 8a b
6 7 2 8 4 9 6
=
2 3 5 3 2 3 2 2 3 5 2 3 5 2 3 5 3
( a − bc ) = ( a ) − ⋅( a ) ⋅ bc + ⋅ a ⋅( bc ) − ( bc ) =
3 2 3 2 2 2 3 5 2 2 3 2 5 2 3 3
= 3 ⋅( a ) −3⋅3 ⋅2⋅( a ) ⋅b ⋅ c + 9⋅a
⋅2
⋅( b ) ⋅( c ) − 2 ⋅( b )
38) 3 2 3 3 3 2 3 3 2 2
2⋅3 2⋅2 3 5 2 3⋅2 5⋅2 3⋅2 5⋅2
= 27⋅ −3⋅9⋅2⋅ ⋅ ⋅ + 9⋅4⋅ ⋅ ⋅ − 8⋅ ⋅ =
a a b c a b c b c
= 27a − 54a b c + 36a b c −8b c
6 4 3 5 2 6 10 6 10
( c )
⋅ =
2 5 2
2 3 4 3 2 3 2 2 3 4 2 3 4 2 3 4 3
( x + yz) = ( x) + ⋅( x) ⋅ yz + ⋅ x⋅ ( yz) + ( yz)
=
= 5 ⋅ ( x ) + 3⋅5 ⋅( x ) . y ⋅ z + 15⋅x ⋅( y ) ⋅ ( z ) + ( y ) ⋅( z )
39) 5 5 3 5 3 5
3 2 3 2 2 2 3 4 2 3 2 4 2 3 3 4 3
2⋅3 2⋅2 3 4 2 3⋅2 4⋅2 3⋅3 4⋅3
= 125⋅ x + 3⋅25⋅x ⋅y ⋅ z + 15⋅x ⋅y ⋅ z + y ⋅ z =
= 125x + 75x y z + 15x y z + y z
6 4 3 4 2 6 8 9 12
=
3 3 2
1 3 ⎛1 ⎞ ⎛1⎞ ⎛1⎞
1 2 3
−
( )
40) x + x = ⎜ + x⎟ = ⎜ ⎟ + 3⋅⎜ ⎟ ⋅ x+ 3⋅ ⋅ x + x =
⎝ x ⎠ ⎝ x⎠ ⎝ x⎠
x
1 1
x x
3 2 2
3
= + 3⋅ ⋅ x+ 3⋅ + x =
3 2
1
1⋅
x
= 3
3
x
+ ⋅ x ⋅ x
x
x
+ 3⋅
x⋅
x
x
1 3
= + + 3x
+ x
3
x x
3
3
+ x =
10
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2 3
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+ 3AB −B
3 3 2
2 3 ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 1 2 3
2 2 2 2
−
( )
41) x − x = ⎜ − x⎟ = ⎜ ⎟ −3⋅⎜ ⎟ ⋅ x+ 3⋅ ⋅x − x = → x
⎝ x ⎠ ⎝ x ⎠ ⎝ x ⎠ x
3
= 1
( x )
1
− 3⋅ ⋅ x + 3⋅
x
( x )
2 3 2 2
2 2
3
2
x
3
− =
1 1
3
= − 3⋅ ⋅ x + 3⋅1− x =
23 ⋅
22 ⋅
x x
1 x
3
x 1⋅
x
= − 3⋅ + 3 − x = → 3⋅ = 3⋅
6 4 4
x x x x⋅
x
1 3
= + + 3 + x
6 3
x x
x
−2
1
=
2
x
1 3
= 3⋅ =
x x
3 3 3
42)
3 2 3 ⎛ 1 2⎞
x + x = ⎜ + x =
3 ⎟
⎝ x ⎠
−
( )
( x ) ( x )
3
3 2
2 2 2 2 3
3 x 3 x x
3 3 3
⎛ 1 ⎞ ⎛ 1 ⎞ 1
= ⎜ ⎟ + ⋅⎜ ⎟ ⋅ + ⋅ ⋅ ( ) + ( ) =
⎝ x ⎠ ⎝ x ⎠ x
3 2
1 1 2 3 2⋅2 2⋅3
3 x x x
3 3 3 2
3
= + ⋅ ⋅ + ⋅ + =
x
2 4 4
1 31 ⋅⋅x
3⋅x
6
3⋅x
= + + + x = → vidi : pod
33 ⋅ 32 ⋅
() 1
3 3
x x x
x
+
1 3⋅x 3⋅x 6
3⋅x
= + + + x =
→ vidi : pod ( 2)
9 6 3 6
x x x x
2 3 1
1 3⋅x 3⋅x ⋅x
6
= + + + x =
9 2+
4 3
x x x
2
1 3⋅
x
= +
9 2
x x ⋅x
2 3 1 2
4
3
3⋅
x ⋅x
+
3
x
1 3
= + + 3x
+ x
9 4
x x
6
1
x
6
+ =
() 1
( 2)
3⋅x 3⋅x 3⋅x ⋅x 3⋅
x
= = =
3 3 3
x x x x
4 3+
1 3 1 1 3
2 2 2 1 2
3⋅x 3⋅x 3⋅x 3⋅
x
= = =
x x x x ⋅x
6 2+
4 2 4 2
⋅
1
x
3
1
4
⋅x
1
1
31 ⋅⋅x
= = 3x
1
31 ⋅ 3
= =
4 4
1⋅
x x
11
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2 3
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+ 3AB −B
m n m m n m n n
( ) ( ) ( ) ( ) ( )
3 3 2 2 3
− = − ⋅ ⋅ + ⋅ ⋅ + =
43) 2 2 2 3 2 2 3 2 2 2
m⋅3 m⋅2 n m n⋅2 n⋅3
= − ⋅ ⋅ + ⋅ ⋅ + =
2 3 2 2 3 2 2 2
= −
3m
2m n m 2n 3n
2 3⋅ 2 ⋅ 2 + 3⋅ 2 ⋅ 2 + 2 =
3m 2m+ n m+
2n 3n
= 2 −3⋅ 2 + 3⋅ 2 + 2 = → možemo ostaviti u ovom obliku ili nastaviti dalje:
2 3 2 3 2 2
3⋅ m 2m+ n m+ 2n 3⋅n
= − ⋅ + ⋅ + =
3 2 + + 2 3
( 2 ) 3 2 3 2 ( 2 )
m
= 8 −3⋅2
primjeni primjeni
a ⋅ a = a a ⋅ a = a
m n m+ n m n m+
n
m m n m n n
= − ⋅ + ⋅ + =
+ 32 ⋅ + 8
2m+ n m+
2n n
n n 3 n 3 n 2 n n n 2 n 3
n
( − ) = ( ) − ⋅( ) ⋅ + ⋅ ⋅( ) − ( ) = → vidi : pod ()( 1 )
44) 3 2 3 3 3 2 3 3 2 2 3
n⋅3 n⋅2 n n n⋅2 n⋅3
= 3 −3⋅3 ⋅ 2 + 3⋅3 ⋅2 − 2 =
3⋅n 2⋅n n n 2⋅n 3⋅n
= 3 −3⋅3 ⋅ 2 + 3⋅3 ⋅2 − 2 =
3 2 2 3
( 3 ) 3 ( 3 ) 2 3 3 ( 2 ) ( 2 )
n n n n n n
= − ⋅ ⋅ + ⋅ ⋅ − =
n n n n n n
= 27 −3⋅ 9
⋅ 2 + 3⋅3 ⋅4 − 8 =
n
n
( )
( )
a ⋅ b = ab
n
n
n
= 27 −3⋅ 9⋅ 2 + 3⋅ 3⋅4 − 8 =
n
( )
= 27 −3⋅ 18 + 3⋅12 −8
n n n n
n
3
n n 3 n 3 n 2 n n n 2 n 3
( + ) = ( ) + ⋅( ) ⋅ + ⋅ ⋅ ( ) + ( ) = → vidi : pod ( 2)
( ) 3
45) 2 3 2 3 2 3 3 2 3 3 2
n⋅3 n⋅2 n n n⋅2 n⋅3
= 2 + 3⋅2 ⋅ 3 + 3⋅2 ⋅ 3 + 3 =
3⋅n 2⋅n n n 2⋅n 3⋅n
= 2 + 3⋅2 ⋅ 3 + 3⋅2 ⋅ 3 + 3 =
3 2 2 3
( 2 ) 3 ( 2 ) 3 3 2 ( 3 ) ( 3 )
n n n n n n
= + ⋅ ⋅ + ⋅ ⋅ + =
n n n n n n
= 8 + 3⋅4 ⋅ 3 + 3⋅2 ⋅ 9 + 27 =
n
n
= 8 + 3⋅( 4⋅ 3)
+ 3⋅( )
= 8 + 3⋅ 12 + 3⋅ 18 + 27
n n n n
n
n
29 ⋅ + 27 =
n
() ( ) 3 n⋅ 3 3 ⋅n ( 3 n n n
1 3 3 3 3 ) 27 ( 2) ( 2 ) 3 n⋅ 2 3 2
3 ⋅n
= = = = = = = (
2 = 8
3n n
)
n
12
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja + kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2 3
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+ 3AB −B
n n n n n n n n
( a b ) ( a ) ( a ) b a ( b ) ( b )
46) 3 3
3 3 2 2 3
+ = + ⋅ ⋅ + ⋅ ⋅ + =
n⋅3 n⋅2 b n n⋅2 n⋅3
= a + 3⋅a ⋅ b + 3⋅a ⋅ b + b =
3n
= a +
3a b + 3a b + b
2n n n 2n 3n
n n+ n n n+ n n+ n+
( a a ) ( a ) ( a ) a a ( a ) ( a )
1 3 3 2 1 1 2 1 3
− = − ⋅ ⋅ + ⋅ ⋅ − =
47) 3 3
( n+ 1) ⋅ 2 ( n+ 1) ⋅ 3 ( n+ 1) ⋅3 3⋅ ( n+
1)
= a −3⋅a ⋅ a + 3⋅a ⋅a − a = → a = a = a
n⋅3 n⋅ 2 n+ 1 n 3n+
3
3n 2n n+ 1 n 2n+ 2 3n+
3
= a −3⋅a ⋅ a + 3⋅a ⋅a − a = ⎪
3n 2n+ n+ 1 n+ 2n+ 2 3n+
3
= a −3⋅ a + 3⋅a − a = ⎪⎭
= a − 3a + 3a −a
3n 3n+ 1 3n+ 2 3n+
3
⎫
⎬
→ a ⋅ a = a
= a
n 2n+ 2 n+
2n+ 2 3n+
2
n+ n− n+ n+ n− n+ n− n−
( a a ) ( a ) ( a ) a a ( a ) ( a )
1 1 3 1 3 1 2 1 1 1 2 1 3
48) + = + 3⋅ ⋅ + 3⋅ ⋅ + =
( n ) ( n ) n− 1 n+
1 ( n ) ( n )
+ 1 ⋅ 3 + 1 ⋅2 −1 ⋅2 −1 ⋅3
= a + 3⋅a ⋅ a + 3⋅a ⋅ a + a =
3n+ 3 2n+ 2 n− 1 n+ 1 2n−2 3n−3
= + ⋅ ⋅ + ⋅ ⋅ + =
a 3 a a 3 a a a
= a + 3⋅ a + 3⋅ a + a
3n+ 3 2n+ 2+ n− 1 n+ 1+ 2n−2 3n−3
3n+ 3 2n+ n+ 2− 1 n+ 2n+ 1−2 3n−3
= a + 3⋅ a + 3⋅ a + a =
= a + 3a + 3a + a
3n+ 3 3n+ 1 3n−1 3n−3
=
( ) ( )
3 3 3 2 2 3 3 2 2 3
49) a− 1 + a+ 1 = a −3⋅a ⋅ 1+ 3⋅a⋅1 − 1 + a + 3⋅a ⋅ 1+ 3⋅a⋅ 1 + 1 =
3 2 3 2
= a − 3a + 3a− 1+ a + 3a + 3a+ 1=
3 3 2 2
= a + a − 3a + 3a + 3a+ 3a− 1+ 1=
= +
3
2a
6
a
( )
2
= − 3+ 3 a = 0
( a−
)
50) 1
3
3 3 2 2 3 3 2 2 3
−( a+ 1) = a −3⋅a ⋅ 1+ 3⋅a⋅1 −1 − ( a + 3⋅a ⋅ 1+ 3⋅a⋅ 1 + 1 ) =
( 3 3) a
2
( )
3 2 3 2
= a − 3a + 3a−1− a + 3a + 3a+ 1 =
3 2 3 2
= a − 3a + 3a−1−a −3a −3a− 1=
3 3 2 2
= a −a −
3a − 3a + 3a−3a−1− 1=
= 0 = − −
= 0
=− −
2
6a
2
13
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2 3
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+ 3AB −B
3 3 3 2 2 3 3 2 2 3
( x y) ( x y) x x y xy y ( x x y xy y )
51) + − − = + 3 + 3 + − − 3 + 3 − =
= + 3 + 3 + − + 3 −3
3 2 2 3 3 2
2 3
x x y xy y x x y xy
3 3 2 2 2 2 3 3
= x − x + 3x y+ 3x y+ 3xy − 3xy + y + y =
= 0
= + = 0
= + y
= 6xy+
2y
2 3
( 3 3) x y
( 1 1)
2 3
+ y =
( ) ( )
3 3 3 2 2 3 3 2 2 3
52) x+ y + x− y = x + 3x y+ 3xy + y + x − 3x y+ 3xy − y =
3 3 2 2 2 2 3 3
= x + x + 3x y− 3x y+ 3xy + 3xy + y − y =
= 2x
+ 6xy
3 2
= 0 = 0
( x )
3
( ) ( ) (
( ) (
3 3 2 2 3 3 2
53) 2 + 3 −3
x− 1 = 2⋅ x + 3⋅x ⋅ 3+ 3⋅x⋅ 3 + 3 −3⋅ x −3⋅x ⋅ 1+ 3⋅x⋅1
( 2 3) x ( 18 9)
3 2
)
( )
3 2 3 2
= 2⋅ x + 9x + 27x+ 27 −3⋅ x − 3x + 3x− 1 =
= ⋅ + ⋅ + ⋅ + ⋅ − ⋅ − ⋅ − − ⋅
3 2 3 2
2 x 2 9x 2 27x 2 27 3 x 3 3x
3 3
3 2 3 2
= 2x + 18x + 54x+ 54− 3x + 9x − 9x+ 3=
3 3 2 2
= 2 x − 3x + 18x + 9x + 54x−9x
+ 54+ 3 =
= − = +
3 2
=− x + 27x + 45x+
57
x
( 54 9)
= −
x
)
− 1 =
2 3
( )
x−3⋅ − 1 =
3 3 3 2 2 3 3 2
( x ) ( x) ( x x x ) ( x
3 2 2 3
3 ( x 3x 3x 1) 5 ( 8 12x 6x x )
54) 3 + 1 −5 2− = 3⋅ + 3⋅ ⋅ 1+ 3⋅ ⋅ 1 + 1 −5⋅ 2 −3⋅2 ⋅ + 3⋅2⋅x
= ⋅ + + + − ⋅ − + − =
( )
3 2 2
= ⋅ + ⋅ + ⋅ + ⋅ − ⋅ − ⋅ − − ⋅ −
3 x 3 3x 3 3x 3 1 5 8 5 12x 5 6x
3 2 2 3
= + + + − + − + =
3x 9x 9x 3 40 60x 30x 5x
= 3x
+ 5x + 9x − 30x + 9x+ 60x+ 3− 40 =
3 3 2 2
= − + −
3 2
8x 21x 69x
37
2 3
)
− x =
3
( x )
5⋅ − =
14
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+
3AB
−B
2 3
( x y) ( x y)
3 3
55) 3 2 + − 4 + 2 =
3 2 2 3 3 2
( x) ( x) y x y y x x y x ( y)
= 3⋅ ⎡ 2 + 3⋅ 2 ⋅ + 3⋅2 ⋅ + ⎤−4⋅ ⎡ + 3⋅ ⋅ 2 + 3⋅ ⋅ 2
⎣ ⎦ ⎣
( ) (
3 2 2 3 3 2 2 3
( x x y xy y ) ( x x y xy y )
3 2 x 3 4x y 6xy y 4 x 6x y 3 x 4y 2 y
3 3 2 2 3 3 2 2 3 3
= ⋅ + ⋅ ⋅ + + − ⋅ + + ⋅ ⋅ + =
= 3⋅ 8 + 12 + 6 + −4⋅ + 6 + 12 + 8 =
3 2 2 3 3 2 2
38 312 36 3 4 46 412 48
= ⋅ x + ⋅ x y+ ⋅ xy + ⋅y − ⋅x − ⋅ x y− ⋅ xy − ⋅
24x 36x y 18xy 3y 4x 24x y 48xy 32y
3 2 2 3 3 2 2 3
= + + + − − − − =
3 3
24 4 36
= x − x + x
= 20x 12x y−30xy −29y
3 2 2 3
y− 24x y+ 18xy − 48xy + 3y − 32y
=
2 2 2 2 3 3
)
y
3
( 2y)
2 3
=
+
⎤ =
⎦
3 2 3 2 2 3 2 2
( x ) x( x ) x x x x ( x x )
3 2 2
= x − 3x + 3x−1−x⋅ ( x + 4x+ 4)
=
56) −1 − + 2 = −3⋅ ⋅ 1+ 3⋅ ⋅1 −1 − ⋅ + 2⋅ ⋅ 2+ 2 =
= − 3 + 3 −1− ⋅ − ⋅4 − ⋅ 4 =
3 2 2
x x x x x x x x
= − 3 + 3 −1− −4 − 4 =
3 2 1+ 2 1+
1
x x x x x x
3 2 3 2
= x − 3x + 3x−1−x −4x − 4x
=
3 3 2 2
= x −x −3x − 4x + 3x−4x− 1=
2
7x
x 1
=− − −
2 3 2 2 3 2
( ) ( ) ( ) (
2 3 2
x ( x 4x 4) ( x 6x 3 4 x 8)
57) x x−2 − x+ 2 = x⋅ x −2⋅x⋅ 2+ 2 − x + 3⋅x ⋅ 2+ 3⋅x⋅ 2 + 2
= ⋅ − + − + + ⋅ ⋅ + =
( )
2 3 2
= xx ⋅ + x⋅ − 4x+ x⋅4−x−6x −12x− 8 =
1+ 2 1+
1 3 2
= x − 4x + 4x−x −6x −12x− 8 =
3 3 2 2
= x −x −
4x − 6x + 4x−12x− 8 =
( ) x
2
= 0 =−4−6
2
10 8 8
=− x − x−
( 412)
= −
x
2 3
)
=
15
© M I M Sraga 2005
www mim sraga com

M-1-Kub zbroja i kub razlike
Rješenja
+ kompletni postupak
40.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+
3AB
3
−B
3 3 3 3
( x+ y) − ( x− y) − ( x + y ) =
3 2 2 3 3 2 2 3 3 3
= 2⋅ ( x + 3x y+ 3xy + y ) −5⋅( x − 3x y+ 3xy − y ) −x − y =
3 2 2 3 3 2 2
= 2⋅ x + 2⋅ 3 23 2 5 5( 3 ) 53 5(
58) 2 5
xy+ ⋅ xy + ⋅y− ⋅x− ⋅ − xy− ⋅ xy − ⋅ −y
3 2 2 3 3 2 2 3 3 3
= 2x + 6x y+ 6xy + 2y − 5x + 15x y− 15xy + 5y −x − y =
3 3 3 2 2 2 2 3 3 3
= 2x −5x − x + 6x y+ 15x y+ 6xy − 15xy + 2y + 5y − y =
( 2 5 1) 3
( 615) ( 615) ( 251)
= − − x = + x y = − xy = + − y
=−4x + 21x y− 9xy + 6y
3 2 2 3
2 2 3
)
−x− y =
3 3 3
16
© M I M Sraga 2005
www mim sraga com