Math, asked by yatharthjain12p96ey7, 1 year ago

prove that a3(b-c)whole cube+b 3 (c-a)whole cube+c3 (a-b)whole cube=3abc (a-b)(b-c)(c-a)

Answers

Answered by Saadqusaim
5
. (a+b+c)^3 = a^3+b^3+c^3 + 3(a+b)(b+c)(a+c)

2. a^3+b^3+c^3 - 3 a b c = (a+b+c)(a^2+b^2+c^2 - a b - b c - a c)

Also note that (a^2+b^2+c^2-ab-bc-ac)=1/2((a-b)^2+(b-c)^2+(a-c)^2)
Answered by Anonymous
4
Hey\:There

↪ As we know ,
( x – y )³ = ( x – y ) ( x² + y² – 2xy )


↪ Now , substituting this formula in the given eqⁿ

a³(b – c)³ + b³(c – a)³ + c³(a – b)³ = L.H.S.

= [ a(b–c) ]³ + [ b(c–a) ]³ + [ c(a–b) ]³


↪ As we know,
if a + b + c = 0
then a³ + b³ + c³ = 3abc


↪ Now , a + b + c = a(b–c) + b(c–a) + c(a–b)

= ab – ac + bc – ab + ac – bc

= ( ab – ab ) + ( bc – bc ) + ( ac – ac )

= 0 + 0 + 0

= 0


↪ Therefore, [ a(b–c) ]³ + [ b(c–a) ]³ + [ c(a–b) ]³

= 3 [ a(b–c) ] [ b(c–a) ] [ c(a–b)

= 3abc ( a – b ) ( b – c ) ( c – a )

= R.H.S.



Anonymous: I edited my ans.!! take a glance!! : )
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