Solve (2x+3)^3-(x+3)^3+(2x-5)^2

$$( 2 x + 3 ) ^ { 3 } - ( x + 3 ) ^ { 3 } + ( 2 x - 5 ) ^ { 2 }$$

Evaluate

$$7x^{3}+31x^{2}+7x+25$$

Solution Steps

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

$$8x^{3}+36x^{2}+54x+27-\left(x+3\right)^{3}+\left(2x-5\right)^{2}$$

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

$$8x^{3}+36x^{2}+54x+27-\left(x^{3}+9x^{2}+27x+27\right)+\left(2x-5\right)^{2}$$

To find the opposite of x^{3}+9x^{2}+27x+27, find the opposite of each term.

$$8x^{3}+36x^{2}+54x+27-x^{3}-9x^{2}-27x-27+\left(2x-5\right)^{2}$$

Combine 8x^{3} and -x^{3} to get 7x^{3}.

$$7x^{3}+36x^{2}+54x+27-9x^{2}-27x-27+\left(2x-5\right)^{2}$$

Combine 36x^{2} and -9x^{2} to get 27x^{2}.

$$7x^{3}+27x^{2}+54x+27-27x-27+\left(2x-5\right)^{2}$$

Combine 54x and -27x to get 27x.

$$7x^{3}+27x^{2}+27x+27-27+\left(2x-5\right)^{2}$$

Subtract 27 from 27 to get 0.

$$7x^{3}+27x^{2}+27x+\left(2x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$7x^{3}+27x^{2}+27x+4x^{2}-20x+25$$

Combine 27x^{2} and 4x^{2} to get 31x^{2}.

$$7x^{3}+31x^{2}+27x-20x+25$$

Combine 27x and -20x to get 7x.

$$7x^{3}+31x^{2}+7x+25$$

Expand

$$7x^{3}+31x^{2}+7x+25$$

Solution Steps

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

$$8x^{3}+36x^{2}+54x+27-\left(x+3\right)^{3}+\left(2x-5\right)^{2}$$

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

$$8x^{3}+36x^{2}+54x+27-\left(x^{3}+9x^{2}+27x+27\right)+\left(2x-5\right)^{2}$$

To find the opposite of x^{3}+9x^{2}+27x+27, find the opposite of each term.

$$8x^{3}+36x^{2}+54x+27-x^{3}-9x^{2}-27x-27+\left(2x-5\right)^{2}$$

Combine 8x^{3} and -x^{3} to get 7x^{3}.

$$7x^{3}+36x^{2}+54x+27-9x^{2}-27x-27+\left(2x-5\right)^{2}$$

Combine 36x^{2} and -9x^{2} to get 27x^{2}.

$$7x^{3}+27x^{2}+54x+27-27x-27+\left(2x-5\right)^{2}$$

Combine 54x and -27x to get 27x.

$$7x^{3}+27x^{2}+27x+27-27+\left(2x-5\right)^{2}$$

Subtract 27 from 27 to get 0.

$$7x^{3}+27x^{2}+27x+\left(2x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$7x^{3}+27x^{2}+27x+4x^{2}-20x+25$$

Combine 27x^{2} and 4x^{2} to get 31x^{2}.

$$7x^{3}+31x^{2}+27x-20x+25$$

Combine 27x and -20x to get 7x.

$$7x^{3}+31x^{2}+7x+25$$