# SOLVE X6

\$\$\oxeda^3-b^3 = (a- b)(a^2+ ab +b^2)\\ oxeda^3+ b^3 = (a+ b)(a^2- ab +b^2)\$\$

Now lets look at your question.

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\$\$\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)>\\ = \\\$\$ +122377

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 phối 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 mix of parentheses together. Finally, write a plus & then the square of the second term in the first mix of parentheses. So we should have this.....

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

CPhill Apr 6, 2015
#3

0

ty was confused why it was (x^2+y^2)

Guest Apr 6, 2015
#4 +117042
+10

These are 2 factorizations that you are supposed khổng lồ memorized

\$\$\oxeda^3-b^3 = (a- b)(a^2+ ab +b^2)\\ oxeda^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)>\\ = \\\$\$

Melody Apr 7, 2015

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