Solve f(x)=(3x^2+1)^3

$$f ( x ) = ( 3 x ^ { 2 } + 1 ) ^ { 3 }$$

Evaluate

$$\left(3x^{2}+1\right)^{3}$$

Solution Steps

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

$$27\left(x^{2}\right)^{3}+27\left(x^{2}\right)^{2}+9x^{2}+1$$

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

$$27x^{6}+27\left(x^{2}\right)^{2}+9x^{2}+1$$

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

$$27x^{6}+27x^{4}+9x^{2}+1$$

Expand

$$27x^{6}+27x^{4}+9x^{2}+1$$

Solution Steps

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

$$27\left(x^{2}\right)^{3}+27\left(x^{2}\right)^{2}+9x^{2}+1$$

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

$$27x^{6}+27\left(x^{2}\right)^{2}+9x^{2}+1$$

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

$$27x^{6}+27x^{4}+9x^{2}+1$$