the lcm of x³-a³ and (x-a)² is??
Answers
Step-by-step explanation:
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)
Hope this helps you
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Step-by-step explanation:
x³-a³=(x-a)(x²+a²+ax)
(x-a)²=(x-a)(x-a)
(x-a) is common factor
lcm=(x²+a²+ax)(x-a)(x-a)