Solve (x^2-3)^3

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

Expand

$$x^{6}-9x^{4}+27x^{2}-27$$

Solution Steps

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

$$\left(x^{2}\right)^{3}-9\left(x^{2}\right)^{2}+27x^{2}-27$$

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

$$x^{6}-9\left(x^{2}\right)^{2}+27x^{2}-27$$

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

$$x^{6}-9x^{4}+27x^{2}-27$$

Evaluate

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

Solution Steps

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

$$\left(x^{2}\right)^{3}-9\left(x^{2}\right)^{2}+27x^{2}-27$$

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

$$x^{6}-9\left(x^{2}\right)^{2}+27x^{2}-27$$

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

$$x^{6}-9x^{4}+27x^{2}-27$$