Question 399501
{{{(v+p)(v^2-vp+p^2)}}}
The straightforward way to multiply this is to multiply each term of one factor by each term of the other factor. Then add/subtract like terms, if any. With 2 terms in the first factor and 3 terms in the second factor, this makes a total of 6 multiplications. <br>
But there is a much faster way. Your expression matches the sum of cubes pattern, {{{a^3+b^3 = (a+b)(a^2-ab+b^2)}}}, except with v's and p's instead of a's and b's. If you recognize this then the pattern shows you how it multiplies:
{{{(v+p)(v^2-vp+p^2) = v^3+p^3}}}