#4**+10 **

These are 2 factorizations that you are supposed to memorized

$$\\\boxed{a^3-b^3 = (a- b)(a^2+ ab +b^2)}\\\\

\boxed{a^3+ b^3 = (a+ b)(a^2- ab +b^2)}$$

Now lets look at your question.

$$\\x^6+y^6\\\\

= (x^2)^3+(y^2)^3\\\\

= [(x^2)+(y^2)][(x^2)^2-(x^2y^2)+(y^2)^2]\\\\

= [(x^2)+(y^2)][(x^4-(x^2y^2)+(y^4)]\\\\

= [x^2+y^2][x^4-x^2y^2+y^4]\\\\$$

Melody
Apr 7, 2015

#2**+8 **

x^6 + y^6 ..... this is the sum of two cubes

First......take the cube root of both things = x^2 and y^2 write them in parentheses with a plus between them

Then...in the next set of parentheses, write the square of the first thing above, then write the opposite sign of the one in the first set of parentheses. Then multiply the two terms in the first set of parentheses together. Finally, write a plus and then the square of the second term in the first set of parentheses. So we should have this.....

(x ^2 + y^2) (x^4 - x^2*y^2 + y^4)

CPhill
Apr 6, 2015

#4**+10 **

Best Answer

These are 2 factorizations that you are supposed to memorized

$$\\\boxed{a^3-b^3 = (a- b)(a^2+ ab +b^2)}\\\\

\boxed{a^3+ b^3 = (a+ b)(a^2- ab +b^2)}$$

Now lets look at your question.

$$\\x^6+y^6\\\\

= (x^2)^3+(y^2)^3\\\\

= [(x^2)+(y^2)][(x^2)^2-(x^2y^2)+(y^2)^2]\\\\

= [(x^2)+(y^2)][(x^4-(x^2y^2)+(y^4)]\\\\

= [x^2+y^2][x^4-x^2y^2+y^4]\\\\$$

Melody
Apr 7, 2015