Question

X3-a3=(x-a)(x2-xa+a2)

Answer 1

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DHANA L. asked • 10/21/16

The L.C.M of (x^3 - a^3) and (x - a)^2 is

explain the answer.......... please

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Kenneth S. answered • 10/21/16

Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018

Factor each and use each distinct factor as little as possible but enough times to contain each factor that either of the two expressions contains. This requires knowing the DIFFERENCE OF CUBES.

x3-a3 = (x-a)(x2+ax+a2) & therefore the LCM is (x-a)2(x2+ax+a2).

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Peter G. answered • 10/21/16

Success in math and English; Math/Logic Master's; 99th-percentile

First, x3-a3 = (x-a)(x2+ax+a2). This can be verified by multiplying out the expression on the right. (If you don't know the factorization on the right, you can find it by noticing that x3-a3 has "a" as a root, and therefore x-a must divide it. Now use polynomial long division to get the expression on the right.)

The challenge is to check that (x-a) and (x2+ax+a2) are relatively prime. By the Euclidean algorithm for polynomials (polynomial long division with a treated as a constant that can be inverted if necessary) the GCD of the two is a constant multiple of the remainder, 3a3, from dividing (x2+ax+a2) by (x-a). Since that remainder has degree 0, but is itself non-zero, the two are relatively prime. Another way to see that is to note that (x-a) is irreducible since it is degree 1, so if it does not divide (x2+ax+a2) then they are relatively prime. For it to divide it, "a" must be a root of (x2+ax+a2). But plugging in "a" gives a non-zero result. Therefore they are relatively prime.

Therefore we can combine factors to get the LCM: (x-a)(x-a)(x2+ax+a2)

Answer 2

To prove that: $x^3$ - $a^3$ = (x - a)($x^2$ + xa + $a^2$).

Solution:

R.H.S. = (x - a)($x^2$+ xa + $a^2$)

= x($x^2$+ xa + $a^2$) - a($x^2$+ xa + $a^2$)

= $x^3$ + $x^2$a + x $a^2$ - $x^2$a - x $a^2$ - $a^3$

= $x^3$ + ($x^2$a - $x^2$a) + (x $a^2$ - x $a^2$ ) - $a^3$

= $x^3$ + 0 + 0 - $a^3$

= $x^3$ - $a^3$

= L.H.S., proved

∴ $x^3$ - $a^3$ = (x - a)($x^2$ + xa + $a^2$), proved.

Thus, $x^3$ - $a^3$ = (x - a)($x^2$ + xa + $a^2$), proved.