Math, asked by sahil359185, 5 months ago

Prove that a3+b3+c3

-3abc=1/2(a+b+c){(a-b)2+(b-c)2+(c-a)2

}​

Answers

Answered by Anonymous
3

Answer:

a^3 + b^3 + c^3 - 3abc = (a+b+c) (a^2 + b^2 + c^2 - ab - bc -ca)

So if a+b+c = 0  

then a^3 + b^3 + c^3 - 3abc = 0

So a^3 + b^3 + c^3 = 3abc

Hope it helps and pls mark as the best answer.

Step-by-step explanation:

Answered by avinashsahni0951
3

Prove that (a + b + c) 3 – a

3 – b

3 – c

3 = 3(a + b ) (b + c ) (c + a).

Answer 9:

LHS = (a + b + c) 3 – a

3 – b

3 – c

3

= [(a + b + c) 3 – a

3

] – [b

3 + c

3

]

= [{(a + b + c) – a}{(a + b + c)

2 + (a + b + c)a + a

2}]– [(b + c)(b

2 – bc + c

2

)]

[∵ a

3 – b

3 = (a – b)(a

2 + ab + b2

) and a

3 + b

3 = (a + b)(a

2 – ab + b2

)]

= (b + c){(a

2 + b2 + c

2 + 2ab + 2bc + 2ca) + (a

2 + ab + ac) + a

2}– [(b + c)(b2 –

bc + c2

)]

= (b + c)(3a

2 + b2+ c

2+ 3ab + 2bc + 3ca) – [(b + c)(b

2 – bc + c

2

)]

= (b + c)[(3a

2 + b2+ c

2+ 3ab + 2bc + 3ca) – (b

2 – bc + c

2

)]

= (b + c)[3a

2 + 3ab + 3bc + 3ca]

= 3(b + c)[a

2 + ab + bc + ca]

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

= 3(a + b) (b + c)(a + c)

= 3(a + b) (b + c)(c + a)

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