Solve left(x+3right)^3-left(x-4right)^3=21x^2+7

$${ \left(x+3 \right) }^{ 3 } - { \left(x-4 \right) }^{ 3 } =21 { x }^{ 2 } +7$$

Solve for x

$$x=4$$

Steps for Solving Linear Equation

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

$$x^{3}+9x^{2}+27x+27-\left(x-4\right)^{3}=21x^{2}+7$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-4\right)^{3}.

$$x^{3}+9x^{2}+27x+27-\left(x^{3}-12x^{2}+48x-64\right)=21x^{2}+7$$

To find the opposite of x^{3}-12x^{2}+48x-64, find the opposite of each term.

$$x^{3}+9x^{2}+27x+27-x^{3}+12x^{2}-48x+64=21x^{2}+7$$

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

$$9x^{2}+27x+27+12x^{2}-48x+64=21x^{2}+7$$

Combine 9x^{2} and 12x^{2} to get 21x^{2}.

$$21x^{2}+27x+27-48x+64=21x^{2}+7$$

Combine 27x and -48x to get -21x.

$$21x^{2}-21x+27+64=21x^{2}+7$$

Add 27 and 64 to get 91.

$$21x^{2}-21x+91=21x^{2}+7$$

Subtract 21x^{2} from both sides.

$$21x^{2}-21x+91-21x^{2}=7$$

Combine 21x^{2} and -21x^{2} to get 0.

$$-21x+91=7$$

Subtract 91 from both sides.

$$-21x=7-91$$

Subtract 91 from 7 to get -84.

$$-21x=-84$$

Divide both sides by -21.

$$x=\frac{-84}{-21}$$

Divide -84 by -21 to get 4.

$$x=4$$