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

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

Evaluate

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

Solution Steps

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

$$8\left(x^{2}\right)^{3}-36\left(x^{2}\right)^{2}x+54x^{2}x^{2}-27x^{3}$$

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

$$8x^{6}-36\left(x^{2}\right)^{2}x+54x^{2}x^{2}-27x^{3}$$

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

$$8x^{6}-36x^{4}x+54x^{2}x^{2}-27x^{3}$$

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

$$8x^{6}-36x^{5}+54x^{2}x^{2}-27x^{3}$$

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

$$8x^{6}-36x^{5}+54x^{4}-27x^{3}$$

Expand

$$8x^{6}-36x^{5}+54x^{4}-27x^{3}$$

Solution Steps

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

$$8\left(x^{2}\right)^{3}-36\left(x^{2}\right)^{2}x+54x^{2}x^{2}-27x^{3}$$

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

$$8x^{6}-36\left(x^{2}\right)^{2}x+54x^{2}x^{2}-27x^{3}$$

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

$$8x^{6}-36x^{4}x+54x^{2}x^{2}-27x^{3}$$

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

$$8x^{6}-36x^{5}+54x^{2}x^{2}-27x^{3}$$

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

$$8x^{6}-36x^{5}+54x^{4}-27x^{3}$$