Solve (6x^2-3x)^3

$$( 6 x ^ { 2 } - 3 x ) ^ { 3 }$$

Evaluate

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

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(6x^{2}-3x\right)^{3}.

$$216\left(x^{2}\right)^{3}-324\left(x^{2}\right)^{2}x+162x^{2}x^{2}-27x^{3}$$

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

$$216x^{6}-324\left(x^{2}\right)^{2}x+162x^{2}x^{2}-27x^{3}$$

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

$$216x^{6}-324x^{4}x+162x^{2}x^{2}-27x^{3}$$

To multiply powers of the same base, add their exponents. Add 4 and 1 to get 5.

$$216x^{6}-324x^{5}+162x^{2}x^{2}-27x^{3}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$216x^{6}-324x^{5}+162x^{4}-27x^{3}$$

Expand

$$216x^{6}-324x^{5}+162x^{4}-27x^{3}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(6x^{2}-3x\right)^{3}.

$$216\left(x^{2}\right)^{3}-324\left(x^{2}\right)^{2}x+162x^{2}x^{2}-27x^{3}$$

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

$$216x^{6}-324\left(x^{2}\right)^{2}x+162x^{2}x^{2}-27x^{3}$$

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

$$216x^{6}-324x^{4}x+162x^{2}x^{2}-27x^{3}$$

To multiply powers of the same base, add their exponents. Add 4 and 1 to get 5.

$$216x^{6}-324x^{5}+162x^{2}x^{2}-27x^{3}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$216x^{6}-324x^{5}+162x^{4}-27x^{3}$$