Question

Przekształć wielomiany , stosując wzory na sume lub różnicę sześcianów
a) 8x3-27
b) 125+x3
c)(x-1)3-(x+1)3
d) (2x-3)3+(x+3)3
e) (5x-3)3-(3x+5)3

Answer 1

A) 8x3-27
(2x)³-3³
a = 2x
b = 3
a³-b³ = (a-b)(a²+ab+b²)
(2x-3)[(2x)²+2x*3+3²] = (2x-3)(4x²+6x+9)

b) 125+x³ = 5³+x³
a = 5
b = x
a³+b³ = (a+b)(a²-ab+b²)
(5+x)(5²-5x+x²) = (5+x)(25-5x+x²)

c)(x-1)3-(x+1)3
(x-1)³-(x+1)³ =
a = x-1
b = x+1
wzór: a³-b³ = (a-b)(a²+ab+b²) czyli:
(x-1-x-1)[(x-1)²+(x-1)(x+1)+(x+1)²] = -2(x²-2x+1+x²-1+x²+2x+1) =
-2(3x²+1) = -6x²-2

d) (2x-3)3+(x+3)3
(2x-3)³+(x+3)³
a³+b³ = (a+b)(a²-ab+b²)
a = 2x-3
b = x+3
(2x-3+x+3)[(2x-3)²-(2x-3)(x+3)+(x+3)²] =
3x(4x²-12x+9-2x²-6x+3x+9+x²+6x+9) = 3x(3x²-9x+27) =
9x³-27x²+81x

e) (5x-3)3-(3x+5)3
(5x-3)³-(3x+5)³
a = 5x-3
b = 3x+5
(a-b)(a²+ab+b²)
(5x-3-3x-5)[(5x-3)²+(5x-3)(3x+5)+(3x+5)²] = (2x-8)(25x²-30x+9+15x²+25x-9x-15+9x²+30x+25) = (2x-8)(49x²+16x+19) =
98x³+32x²+38x-392x²-128x-152 = 98x³-360x²-90x-152

Answer 2

WZORY
a³- b³= (a-b)(a²+ ab+ b²)
a³+ b³= (a+b)(a²- ab+ b²)

a) 8x³-27= 2³x³- 3³= (2x)³- 3³= (2x- 3)[(2x)²+ 2x*3+ 3²]=
(2x- 3)(4x²+ 6x+ 9)

b) 125+ x³= 5³+ x³= (5+ x)(5²-5x+ x²)= (5+ x)(25- 5x+ x²)

c)(x-1)³-(x+1)³= [(x-1)-(x+1)][(x-1)²+ (x-1)(x+1)+ (x+1)²]

[(x-1)-(x+1)][(x-1)²+ (x-1)(x+1)+ (x+1)²]=(-2)*[x²-2x+1+x²-1+x²+2x+1]= (-2)*(3x²+ 1)

d) (2x-3)³+(x+3)³=
[(2x-3)+(x+3)][(2x-3)²- (2x-3)(x+3)+ (x+3)²]

[(2x-3)+(x+3)][(2x-3)²- (2x-3)(x+3)+ (x+3)²]=
(3x)*[4x²-12x+9-2x²-6x+3x+9+x²+6x+9]= (3x)*(3x²-9x+27)

e) (5x-3)³-(3x+5)³=
[(5x-3)-(3x+5)][(5x-3)²+ (5x-3)(3x+5)+ (3x+5)²]

[(5x-3)-(3x+5)][(5x-3)²+ (5x-3)(3x+5)+ (3x+5)²]=
(2x+2)*[25x²-30x+9+15x²+25x-9x-15+9x²+30x+25]=
(2x+2)(49x²+16x+19)