Sagot :
Step-by-step explanation:
The sum or difference of two cubes can be factored into a product of a binomial times a trinomial.
That is, x3+y3=(x+y)(x2−xy+y2)" role="presentation" style="display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">x3+y3=(x+y)(x2−xy+y2)x3+y3=(x+y)(x2−xy+y2) and x3−y3=(x−y)(x2+xy+y2)" role="presentation" style="display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">x3−y3=(x−y)(x2+xy+y2)x3−y3=(x−y)(x2+xy+y2) .
A mnemonic for the signs of the factorization is the word "SOAP", the letters stand for "Same sign" as in the middle of the original expression, "Opposite sign", and "Always Positive".
That is, x3±y3=(x[Same  sign]y)(x2[Opposite  sign]xy[Always  Positive]y2)" role="presentation" style="display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">x3±y3=(x[Same sign]y)(x2[Opposite sign]xy[Always Positive]y2)x3±y3=(x[Same sign]y)(x2[Opposite sign]xy[Always Positive]y2)
Example 1:
Factor 27p3+q3" role="presentation" style="display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">27p3+q327p3+q3 .
Try to write each of the terms as a cube of an expression.
27p3+q3=(3p)3+(q)3" role="presentation" style="display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">27p3+q3=(3p)3+(q)327p3+q3=(3p)3+(q)3
Use the factorization of sum of cubes to rewrite.
27p3+q3=(3p)3+(q)3                       =(3p+q)((3p)2−3pq+q2)                       =(3p+q)(9p2−3pq+q2)" role="presentation" style="display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">27p3+q3=(3p)3+(q)3 =(3p+q)((3p)2−3pq+q2) =(3p+q)(9