Solve (x-3)^3+(x+3)^3geq2(x+1)(x^2-x+1)+50x

$$( x - 3 ) ^ { 3 } + ( x + 3 ) ^ { 3 } \geq 2 ( x + 1 ) ( x ^ { 2 } - x + 1 ) + 50 x$$

Solve for x

$$x\geq \frac{1}{2}$$

Solution Steps

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

$$x^{3}-9x^{2}+27x-27+\left(x+3\right)^{3}\geq 2\left(x+1\right)\left(x^{2}-x+1\right)+50x$$

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

$$x^{3}-9x^{2}+27x-27+x^{3}+9x^{2}+27x+27\geq 2\left(x+1\right)\left(x^{2}-x+1\right)+50x$$

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

$$2x^{3}-9x^{2}+27x-27+9x^{2}+27x+27\geq 2\left(x+1\right)\left(x^{2}-x+1\right)+50x$$

Combine -9x^{2} and 9x^{2} to get 0.

$$2x^{3}+27x-27+27x+27\geq 2\left(x+1\right)\left(x^{2}-x+1\right)+50x$$

Combine 27x and 27x to get 54x.

$$2x^{3}+54x-27+27\geq 2\left(x+1\right)\left(x^{2}-x+1\right)+50x$$

Add -27 and 27 to get 0.

$$2x^{3}+54x\geq 2\left(x+1\right)\left(x^{2}-x+1\right)+50x$$

Use the distributive property to multiply 2 by x+1.

$$2x^{3}+54x\geq \left(2x+2\right)\left(x^{2}-x+1\right)+50x$$

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

$$2x^{3}+54x\geq 2x^{3}+2+50x$$

Subtract 2x^{3} from both sides.

$$2x^{3}+54x-2x^{3}\geq 2+50x$$

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

$$54x\geq 2+50x$$

Subtract 50x from both sides.

$$54x-50x\geq 2$$

Combine 54x and -50x to get 4x.

$$4x\geq 2$$

Divide both sides by 4. Since 4 is positive, the inequality direction remains the same.

$$x\geq \frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$x\geq \frac{1}{2}$$