Factorize

$\displaystyle

= y^6-125 = (y^3)^2 - (5)^3

$

can i factorize it more? thanksss

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- Aug 19th 2008, 09:10 AMjvignaciofactorizing expression
Factorize

$\displaystyle

= y^6-125 = (y^3)^2 - (5)^3

$

can i factorize it more? thanksss - Aug 19th 2008, 09:48 AMChop Suey
This is difference between two cubes:

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

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

===============

$\displaystyle y^6 - 125 = (y^2)^3 - (5)^3$

Can you do it now? - Aug 19th 2008, 09:59 AMSoroban
Hello, jvignacio

Quote:

Factorize: .$\displaystyle y^6-125$

$\displaystyle (y^3)^2 - (5)^3$ . . . . This is**not**factored!

For example -- Factor: .$\displaystyle 6x + 21$

"Factor" means to express it as a**product**of two or more expressins.

. . Answer: .$\displaystyle 3 \times (2x + 7)$

This is**not**factored: .$\displaystyle (2)(3)(x) + (3)(7)$

Certainly,__parts__of it are factored, but it is still a**sum**of two expressions.

We have: .$\displaystyle y^6 - 125 \;=\;(y^2)^3 - (5)^3$ . . . difference of cubes!

. . You are expected to know that: .$\displaystyle a^3 -b^3 \:=\:(a-b)(a^2 + ab +b^2)$

Therefore: . $\displaystyle (y^2)^3 - (5)^3 \;=\;(y^2 - 5)(y^4 + 5y^2 + 25)$