Solve (x-1)(x^2+x+1)-9-2xleq(x-1)^3+x(3x-2)

$$( x - 1 ) ( x ^ { 2 } + x + 1 ) - 9 - 2 x \leq ( x - 1 ) ^ { 3 } + x ( 3 x - 2 )$$

Solve for x

$$x\geq -3$$

Solution Steps

Use the distributive property to multiply x-1 by x^{2}+x+1 and combine like terms.

$$x^{3}-1-9-2x\leq \left(x-1\right)^{3}+x\left(3x-2\right)$$

Subtract 9 from -1 to get -10.

$$x^{3}-10-2x\leq \left(x-1\right)^{3}+x\left(3x-2\right)$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-1\right)^{3}.

$$x^{3}-10-2x\leq x^{3}-3x^{2}+3x-1+x\left(3x-2\right)$$

Use the distributive property to multiply x by 3x-2.

$$x^{3}-10-2x\leq x^{3}-3x^{2}+3x-1+3x^{2}-2x$$

Combine -3x^{2} and 3x^{2} to get 0.

$$x^{3}-10-2x\leq x^{3}+3x-1-2x$$

Combine 3x and -2x to get x.

$$x^{3}-10-2x\leq x^{3}+x-1$$

Subtract x^{3} from both sides.

$$x^{3}-10-2x-x^{3}\leq x-1$$

Combine x^{3} and -x^{3} to get 0.

$$-10-2x\leq x-1$$

Subtract x from both sides.

$$-10-2x-x\leq -1$$

Combine -2x and -x to get -3x.

$$-10-3x\leq -1$$

Add 10 to both sides.

$$-3x\leq -1+10$$

Add -1 and 10 to get 9.

$$-3x\leq 9$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

$$x\geq \frac{9}{-3}$$

Divide 9 by -3 to get -3.

$$x\geq -3$$