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

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

Evaluate

$$\left(x^{3}-9\right)\left(x^{3}+6x+9\right)$$

Solution Steps

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

$$\left(x^{3}\right)^{2}+6x^{3}x+9x^{2}-\left(3x+9\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$x^{6}+6x^{3}x+9x^{2}-\left(3x+9\right)^{2}$$

To multiply powers of the same base, add their exponents. Add 3 and 1 to get 4.

$$x^{6}+6x^{4}+9x^{2}-\left(3x+9\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+9\right)^{2}.

$$x^{6}+6x^{4}+9x^{2}-\left(9x^{2}+54x+81\right)$$

To find the opposite of 9x^{2}+54x+81, find the opposite of each term.

$$x^{6}+6x^{4}+9x^{2}-9x^{2}-54x-81$$

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

$$x^{6}+6x^{4}-54x-81$$

Expand

$$x^{6}+6x^{4}-54x-81$$

Solution Steps

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

$$\left(x^{3}\right)^{2}+6x^{3}x+9x^{2}-\left(3x+9\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$x^{6}+6x^{3}x+9x^{2}-\left(3x+9\right)^{2}$$

To multiply powers of the same base, add their exponents. Add 3 and 1 to get 4.

$$x^{6}+6x^{4}+9x^{2}-\left(3x+9\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+9\right)^{2}.

$$x^{6}+6x^{4}+9x^{2}-\left(9x^{2}+54x+81\right)$$

To find the opposite of 9x^{2}+54x+81, find the opposite of each term.

$$x^{6}+6x^{4}+9x^{2}-9x^{2}-54x-81$$

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

$$x^{6}+6x^{4}-54x-81$$