Sum of Two Cubes | MathHelp.com

In this example, we’re asked to factor 125p cubed + 1. Notice that 125p cubed is a perfect cube, because 125 is 5 times 5 times 5, or 5 cubed, and 1 is a perfect cube, because 1 is 1 times 1 times 1, or 1 cubed. So we have the sum of two cubes. To factor the sum of two cubes, we use the following formula: a cubed + b cubed can be factored as parentheses a + b times parentheses a squared – ab + b squared. And in this problem, since a cubed is represented by 125p cubed, the value of a is 5p, and since b cubed is represented by 1, the value of b is 1. Notice that the value of a, 5p, has both a coefficient and a variable, so we need to use parentheses around the 5p when we plug it into the formula, in order to keep the 5 and the p together. And let’s go ahead and use parentheses around our x as well, in order to be consistent. So we have bracket parentheses 5p + parentheses 1 closed bracket times bracket parentheses 5p squared + parentheses 5p times parentheses 1 + parentheses 1 squared closed bracket. Notice that we change the parentheses in the formula to brackets so that we’re not dealing with too many sets of parentheses. Next, simplifying inside of the second set of brackets, parentheses 5p squared is 5p times 5p, or 25p squared, 5p times 1 is 5p, so we have – 5p, and parentheses 1 squared is 1, so we have plus 1. And changing the brackets back to parentheses, we have parentheses 5p + 1 times parentheses 25p squared – 5p + 1, which is our final answer.

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PROFF for( a+b)(a^2-ab+b^2)=a^3+b^3