Question

15 points!

factorise a³(b-c)³+b³(c-a)³+c³(a-b)³

note: if u sum up the values in the brackets they will cancel eachother

and maybe it is related to the identity
x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)

class 9 ch-2 maths

first one to give correct answer will be brainliest and explain in paper thnz alot I will pray that u will get 97% + in ur next exam thnx​

Answer 1

Factorise :-

a³ ( b - c )³ + b³ ( c - a )³ + c³ ( a - b )³

= { a ( b - c )}³ + { b ( c - a )}³ + { c ( a - b )}³

[ a ( b - c ) + b( c - a ) + c ( a - b ) = ab - ac + bc - ab + ac - bc = 0 , so we see the value is 0 , therefore we can use the formula ,

x³ + y³ + z³ = 3xyz , where , ( x + y + z ) = 0 ]

Hence ,

{ a ( b - c )}³ + { b ( c - a )}³ + { c ( a - b )}³

= 3 × a ( b - c ) × b( c - a ) × c ( a - b )

= 3abc( b - c ) ( c - a ) ( a - b ) [ • Required Answer ]

Answer 2

Step-by-step explanation:

I hope it may help to you ..