Solve 27x^2-3(3x-1)^2=

$$27 x ^ { 2 } - 3 ( 3 x - 1 ) ^ { 2 } =$$

Evaluate

$$18x-3$$

Solution Steps

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

$$27x^{2}-3\left(9x^{2}-6x+1\right)$$

Use the distributive property to multiply -3 by 9x^{2}-6x+1.

$$27x^{2}-27x^{2}+18x-3$$

Combine 27x^{2} and -27x^{2} to get 0.

$$18x-3$$

Expand

$$18x-3$$

Solution Steps

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

$$27x^{2}-3\left(9x^{2}-6x+1\right)$$

Use the distributive property to multiply -3 by 9x^{2}-6x+1.

$$27x^{2}-27x^{2}+18x-3$$

Combine 27x^{2} and -27x^{2} to get 0.

$$18x-3$$