A) (x-1)^3+ 3x(x-4)+2=x^3
B) x^3 +6x +12x +8=0

I'll solve A) since there is a lot of work to be shown.

Step-by-step explanation:

A)

(x−1)³+3x(x−4)+2=x³

Use binomial theorem to expand  (x−1)³

x³- 3x² + 3x - 1 + 3x(x-4) + 2=x³

Use distributive property on 3x(x-4)

x³- 3x² + 3x - 1 + 3x²-12x + 2=x³

Add like terms

x³ −9x + 1 =x³

Subtract x³ from both sides.

x³ −9x +1 -x³ = 0

Add x³ and -x³

−9x+1=0

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

−9x=−1

Divide both sides by -9.

x= 1/9

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15