Solve (7b^5-b^2)^2

$$( 7 b ^ { 5 } - b ^ { 2 } ) ^ { 2 }$$

Evaluate

$$b^{4}\left(7b^{3}-1\right)^{2}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(7b^{5}-b^{2}\right)^{2}.

$$49\left(b^{5}\right)^{2}-14b^{5}b^{2}+\left(b^{2}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 5 and 2 to get 10.

$$49b^{10}-14b^{5}b^{2}+\left(b^{2}\right)^{2}$$

To multiply powers of the same base, add their exponents. Add 5 and 2 to get 7.

$$49b^{10}-14b^{7}+\left(b^{2}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$49b^{10}-14b^{7}+b^{4}$$

Expand

$$49b^{10}-14b^{7}+b^{4}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(7b^{5}-b^{2}\right)^{2}.

$$49\left(b^{5}\right)^{2}-14b^{5}b^{2}+\left(b^{2}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 5 and 2 to get 10.

$$49b^{10}-14b^{5}b^{2}+\left(b^{2}\right)^{2}$$

To multiply powers of the same base, add their exponents. Add 5 and 2 to get 7.

$$49b^{10}-14b^{7}+\left(b^{2}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$49b^{10}-14b^{7}+b^{4}$$