Question

factorise (x-a)^3+(x-b)^3+(x-c)^3 where x=a+b+c/3​??<br />plz solve this fast​

Answer 1

we have to factorise (x-a)³ +(x-b)³ +(x-c)³ where x=(a + b + c)/3

we know, if (p + q + r) = 0

then, p³ + q³ + r³ = 3pqr

similarly, here p = (x - a), q = (x - b) and r = (x - c)

and p + q + r = { (x - a) + (x - b) + (x - c)} = {3x - (a + b + c)}

= 3 × (a + b + c)/3 - (a + b + c) [ given, x = (a + b + c)/3 ]

= (a + b + c) - (a + b + c)

= 0

hence, (p + q + r) = (x - a) + (x - b) + (x - c) = 0

then we can apply p³ + q³ + r³ = 3pqr

so, (x - a)³ + (x - b)³ + (x - c)³ = 3(x - a)(x - b)(x - c)

hence, answer should be 3(x - a)(x - b)(x - c)