Prove that a3+b3+c3
-3abc=1/2(a+b+c){(a-b)2+(b-c)2+(c-a)2
}
Answers
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:
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)