Solve (1/7x^2-9)^3

$$( 1 / 7 x ^ { 2 } - 9 ) ^ { 3 }$$

Evaluate

$$\frac{\left(x^{2}-63\right)^{3}}{343}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(\frac{1}{7}x^{2}-9\right)^{3}.

$$\frac{1}{343}\left(x^{2}\right)^{3}-\frac{27}{49}\left(x^{2}\right)^{2}+\frac{243}{7}x^{2}-729$$

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

$$\frac{1}{343}x^{6}-\frac{27}{49}\left(x^{2}\right)^{2}+\frac{243}{7}x^{2}-729$$

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

$$\frac{1}{343}x^{6}-\frac{27}{49}x^{4}+\frac{243}{7}x^{2}-729$$

Expand

$$\frac{x^{6}}{343}-\frac{27x^{4}}{49}+\frac{243x^{2}}{7}-729$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(\frac{1}{7}x^{2}-9\right)^{3}.

$$\frac{1}{343}\left(x^{2}\right)^{3}-\frac{27}{49}\left(x^{2}\right)^{2}+\frac{243}{7}x^{2}-729$$

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

$$\frac{1}{343}x^{6}-\frac{27}{49}\left(x^{2}\right)^{2}+\frac{243}{7}x^{2}-729$$

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

$$\frac{1}{343}x^{6}-\frac{27}{49}x^{4}+\frac{243}{7}x^{2}-729$$