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

Question

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

Answer 1

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).

Answer 2

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 3

 Hi Dhana,

You'll want to start by factoring each expression. (x³ - a³) is divisible by (x-a)  

 

(x³ - a³) =  (x-a)(x² + ax + a²)

 

(x-a)² = (x-a)(x-a)

 

Take the highest degree (power/exponent) factor and multiply them together to get the LCM of both

 

(x-a)²*(x² + ax + a²) = (x-a)(x³ -a³) = LCM

 

 

If you think about the method it in simpler terms;

 

it would be like asking what is the LCM(4,14). pretend 4 is (x-a)² then 2 is our (x-a) and 14 is (x³ - a³)

 

14 = 2*7

4 = 2*2

LCM = 7*2*2 = 28