Solve 1000(x+2)^3-1

$$1000 ( x + 2 ) ^ { 3 } - 1$$

Evaluate

$$\left(10x+19\right)\left(100x^{2}+410x+421\right)$$

Solution Steps

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

$$1000\left(x^{3}+6x^{2}+12x+8\right)-1$$

Use the distributive property to multiply 1000 by x^{3}+6x^{2}+12x+8.

$$1000x^{3}+6000x^{2}+12000x+8000-1$$

Subtract 1 from 8000 to get 7999.

$$1000x^{3}+6000x^{2}+12000x+7999$$

Expand

$$1000x^{3}+6000x^{2}+12000x+7999$$

Solution Steps

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

$$1000\left(x^{3}+6x^{2}+12x+8\right)-1$$

Use the distributive property to multiply 1000 by x^{3}+6x^{2}+12x+8.

$$1000x^{3}+6000x^{2}+12000x+8000-1$$

Subtract 1 from 8000 to get 7999.

$$1000x^{3}+6000x^{2}+12000x+7999$$