Solve h) (8x + 8y) ^ 2 - (8x - 8y) ^ 2 - 128xy | Equatix - Math Solver

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Example

3(r+2s)=2t-4

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Question

$$( 8 x + 8 y ) ^ { 2 } - ( 8 x - 8 y ) ^ { 2 } - 128 x y$$

Evaluate

$128xy$

Solution Steps

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(8x+8y\right)^{2}$.

$$64x^{2}+128xy+64y^{2}-\left(8x-8y\right)^{2}-128xy$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(8x-8y\right)^{2}$.

$$64x^{2}+128xy+64y^{2}-\left(64x^{2}-128xy+64y^{2}\right)-128xy$$

To find the opposite of $64x^{2}-128xy+64y^{2}$, find the opposite of each term.

$$64x^{2}+128xy+64y^{2}-64x^{2}+128xy-64y^{2}-128xy$$

Combine $64x^{2}$ and $-64x^{2}$ to get $0$.

$$128xy+64y^{2}+128xy-64y^{2}-128xy$$

Combine $128xy$ and $128xy$ to get $256xy$.

$$256xy+64y^{2}-64y^{2}-128xy$$

Combine $64y^{2}$ and $-64y^{2}$ to get $0$.

$$256xy-128xy$$

Combine $256xy$ and $-128xy$ to get $128xy$.

$$128xy$$

Expand

$128xy$

Solution Steps

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(8x+8y\right)^{2}$.

$$64x^{2}+128xy+64y^{2}-\left(8x-8y\right)^{2}-128xy$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(8x-8y\right)^{2}$.

$$64x^{2}+128xy+64y^{2}-\left(64x^{2}-128xy+64y^{2}\right)-128xy$$

To find the opposite of $64x^{2}-128xy+64y^{2}$, find the opposite of each term.

$$64x^{2}+128xy+64y^{2}-64x^{2}+128xy-64y^{2}-128xy$$

Combine $64x^{2}$ and $-64x^{2}$ to get $0$.

$$128xy+64y^{2}+128xy-64y^{2}-128xy$$

Combine $128xy$ and $128xy$ to get $256xy$.

$$256xy+64y^{2}-64y^{2}-128xy$$

Combine $64y^{2}$ and $-64y^{2}$ to get $0$.

$$256xy-128xy$$

Combine $256xy$ and $-128xy$ to get $128xy$.

$$128xy$$