Question

a)(x+2)³-2(x-1)³+(x-2)(x²+2x+4)
b)(x-√2)(x+√2)-(x+2)(x²-2x+4)
c)(√2x+1)³-3(x-√2)²-(√2x-1)³

Answer 1

a)(x+2)³=x³+3x²·2+3x·4+8=x³+6x²+12x+8
(x-1)³=x³-3x²+3x+1
(x-2)(x²+2x+4)=x³+2x²+4x-2x²-4x-8=x³-8
⇒x³+6x²+12x+8-2(x³-3x²+3x+1)+x³-8=
x³+6x²+12x+8-2x³+6x²-6x-2+x³-8=
12x²+6x-2=0     :2
6x²+3x-1=0
b)(x-√2)(x+√2)-(x+2)(x²-2x+4)=
x²+√2x-√2x-√2√2-(x³-2x²+4x+2x²-4x+8)=
x²-2-x³-8=-x³+x²-10   (-1)=
x³-x²+10

c)(√2x+1)³-3(x-√2)²-(√2x-1)³=
(√2x+1)³=√8x³+3·√4x²+3·√2x+1=2√2x³+6x²+3√2x+1
(x-√2)²=x²-√2x-√2x+2=x²+2
(√2x-1)³=√8x³-3·√4x²+3·√2x+1=2√2x³-6x²+3√2x+1
⇒2√2x³+6x²+3√2x+1-3(x²+2)-(2√2x³-6x²+3√2x+1)=
2√2x³+6x²+3√2x+1-3x²-6-2√2x³+6x²-3√2x-1=
9x²-6=
3(3x²-2)