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)
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. (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)
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)
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↪ 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.
↪ 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:
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