Solve 6 * (a + b) ^ 2 + 5(a ^ 2 - b ^ 2) - 6 * (a + b) ^ 2 | Equatix - Math Solver

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Example

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

a⋅(b-2)=3b

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

a⋅(b-2)=3b

Question

$$6(a+b)^{2}+5(a^{2}-b^{2})-6(a+b)^{2}$$

Evaluate

$5\left(a^{2}-b^{2}\right)$

Solution Steps

Use binomial theorem $\left(p+q\right)^{2}=p^{2}+2pq+q^{2}$ to expand $\left(a+b\right)^{2}$.

$$6\left(a^{2}+2ab+b^{2}\right)+5\left(a^{2}-b^{2}\right)-6\left(a+b\right)^{2}$$

Use the distributive property to multiply $6$ by $a^{2}+2ab+b^{2}$.

$$6a^{2}+12ab+6b^{2}+5\left(a^{2}-b^{2}\right)-6\left(a+b\right)^{2}$$

Use the distributive property to multiply $5$ by $a^{2}-b^{2}$.

$$6a^{2}+12ab+6b^{2}+5a^{2}-5b^{2}-6\left(a+b\right)^{2}$$

Combine $6a^{2}$ and $5a^{2}$ to get $11a^{2}$.

$$11a^{2}+12ab+6b^{2}-5b^{2}-6\left(a+b\right)^{2}$$

Combine $6b^{2}$ and $-5b^{2}$ to get $b^{2}$.

$$11a^{2}+12ab+b^{2}-6\left(a+b\right)^{2}$$

Use binomial theorem $\left(p+q\right)^{2}=p^{2}+2pq+q^{2}$ to expand $\left(a+b\right)^{2}$.

$$11a^{2}+12ab+b^{2}-6\left(a^{2}+2ab+b^{2}\right)$$

Use the distributive property to multiply $-6$ by $a^{2}+2ab+b^{2}$.

$$11a^{2}+12ab+b^{2}-6a^{2}-12ab-6b^{2}$$

Combine $11a^{2}$ and $-6a^{2}$ to get $5a^{2}$.

$$5a^{2}+12ab+b^{2}-12ab-6b^{2}$$

Combine $12ab$ and $-12ab$ to get $0$.

$$5a^{2}+b^{2}-6b^{2}$$

Combine $b^{2}$ and $-6b^{2}$ to get $-5b^{2}$.

$$5a^{2}-5b^{2}$$

Expand

$5a^{2}-5b^{2}$

Solution Steps

Use binomial theorem $\left(p+q\right)^{2}=p^{2}+2pq+q^{2}$ to expand $\left(a+b\right)^{2}$.

$$6\left(a^{2}+2ab+b^{2}\right)+5\left(a^{2}-b^{2}\right)-6\left(a+b\right)^{2}$$

Use the distributive property to multiply $6$ by $a^{2}+2ab+b^{2}$.

$$6a^{2}+12ab+6b^{2}+5\left(a^{2}-b^{2}\right)-6\left(a+b\right)^{2}$$

Use the distributive property to multiply $5$ by $a^{2}-b^{2}$.

$$6a^{2}+12ab+6b^{2}+5a^{2}-5b^{2}-6\left(a+b\right)^{2}$$

Combine $6a^{2}$ and $5a^{2}$ to get $11a^{2}$.

$$11a^{2}+12ab+6b^{2}-5b^{2}-6\left(a+b\right)^{2}$$

Combine $6b^{2}$ and $-5b^{2}$ to get $b^{2}$.

$$11a^{2}+12ab+b^{2}-6\left(a+b\right)^{2}$$

Use binomial theorem $\left(p+q\right)^{2}=p^{2}+2pq+q^{2}$ to expand $\left(a+b\right)^{2}$.

$$11a^{2}+12ab+b^{2}-6\left(a^{2}+2ab+b^{2}\right)$$

Use the distributive property to multiply $-6$ by $a^{2}+2ab+b^{2}$.

$$11a^{2}+12ab+b^{2}-6a^{2}-12ab-6b^{2}$$

Combine $11a^{2}$ and $-6a^{2}$ to get $5a^{2}$.

$$5a^{2}+12ab+b^{2}-12ab-6b^{2}$$

Combine $12ab$ and $-12ab$ to get $0$.

$$5a^{2}+b^{2}-6b^{2}$$

Combine $b^{2}$ and $-6b^{2}$ to get $-5b^{2}$.

$$5a^{2}-5b^{2}$$