Răspuns :
(a+b)³=a³+3a²b+3ab²+b³
(a-b)³=a³-3a²b+3ab²-b³
aplicam formulelel de mai sus
a)(x+2)³=x³+3x²·2+3x·2²+2³=x³+6x²+12x+8
b)(x-5)³=x³-3x²·5+3x·5²-5³=x³-15x²+75x-125
c)(3x+1)³=3³x³+3·3²x²·1+3·3x·1²+1³=27x³+27x²+9x+1
d)(4x-1)³=4³x³-3·4²x²·1+3·4x·1²-1³=64x³-48x²+12x-1
(a-b)³=a³-3a²b+3ab²-b³
aplicam formulelel de mai sus
a)(x+2)³=x³+3x²·2+3x·2²+2³=x³+6x²+12x+8
b)(x-5)³=x³-3x²·5+3x·5²-5³=x³-15x²+75x-125
c)(3x+1)³=3³x³+3·3²x²·1+3·3x·1²+1³=27x³+27x²+9x+1
d)(4x-1)³=4³x³-3·4²x²·1+3·4x·1²-1³=64x³-48x²+12x-1
a. x^3+6x^2+12x+8
b x^3-15x^2+75x-125
c. 27x^3+27x^2+9x+1
d. 64x^3-48x^2+12x+1
b x^3-15x^2+75x-125
c. 27x^3+27x^2+9x+1
d. 64x^3-48x^2+12x+1
