Question

96. Factorise -(x-a)^3– (x-b)^3 – (x-c)^3, if x = (a+b+c)/3

Pls answer....​

Answer 1

Step-by-step explanation:

Let x-a = p, x-b = q and x-c = r

The given expression becomes p^3+q^3+r^3 -3pqr = (p+q+r)(p^2+q^2+r^2-pq-qr-rp)……..(1)

Now p+q+r= x-a+x-b+x-c = 3x-(a+b+c)=3x-3x =0.

So, p+q+r =0. Hence (1) reduces to 0.

Or the given expression = 0.

Answer 2

To factorize the expression -(x-a)^3 - (x-b)^3 - (x-c)^3, we can substitute x with (a+b+c)/3 and simplify it. Let's proceed with the calculations:

Let x = (a+b+c)/3

Substituting x in the expression:

-((a+b+c)/3 - a)^3 - ((a+b+c)/3 - b)^3 - ((a+b+c)/3 - c)^3

Simplifying further:

-((2a+2b+2c-3a)/3)^3 - ((2a+2b+2c-3b)/3)^3 - ((2a+2b+2c-3c)/3)^3

Simplifying the terms inside the parentheses:

-(a-b)^3 - (b-c)^3 - (c-a)^3

Since (a-b)^3 = -(b-a)^3, (b-c)^3 = -(c-b)^3, and (c-a)^3 = -(a-c)^3, we can rewrite the expression as:

-(a-b)^3 + (b-a)^3 - (c-b)^3 + (c-a)^3 - (a-c)^3 + (c-a)^3

Now, we can simplify further:

-(a-b)^3 + (a-b)^3 + (c-b)^3 - (c-a)^3

The (a-b)^3 term cancels out, and we are left with:

(c-b)^3 - (c-a)^3

Using the difference of cubes formula, (c-b)^3 - (c-a)^3 can be factorized as:

((c-b) - (c-a))((c-b)^2 + (c-b)(c-a) + (c-a)^2)

Simplifying the first factor:

(a-b)((c-b)^2 + (c-b)(c-a) + (c-a)^2)

Therefore, the factorized form of -(x-a)^3 - (x-b)^3 - (x-c)^3, when x = (a+b+c)/3, is (a-b)((c-b)^2 + (c-b)(c-a) + (c-a)^2).

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