Math, asked by VARDHAN34621, 10 hours ago

verify a3+b3+c3-3abc=1/2(a+b+c)( a2+b2+c2- p)

Answers

Answered by fjm5856
0

Solution:

For verifying the given, we need to prove that the left hand side (L.H.S) of the equation is equal to right hand side (R.H.S).

Therefore R.H.S,

= \frac{1}{2}\left((a+b+c)\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right)\right) \ldots \ldots \ldots=

2

1

((a+b+c)((a−b)

2

+(b−c)

2

+(c−a)

2

))………

By using the formula (a-b)^{2}=a^{2}+b^{2}+2 a b(a−b)

2

=a

2

+b

2

+2ab

Then, as per the above formula the equation 1 becomes,

=\frac{1}{2}(a+b+c)\left(a^{2}-2 a b+b^{2}\right)+\left(b^{2}-2 b c+c^{2}\right)+\left(c^{2}-2 c a+a^{2}\right)=

2

1

(a+b+c)(a

2

−2ab+b

2

)+(b

2

−2bc+c

2

)+(c

2

−2ca+a

2

)

=\frac{1}{2}(a+b+c) 2\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=

2

1

(a+b+c)2(a

2

+b

2

+c

2

−ab−bc−ca)

=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=(a+b+c)(a

2

+b

2

+c

2

−ab−bc−ca)

\begin{gathered}\begin{aligned}=& a^{3}+a b^{2}+a c^{2}-b a^{2}-a b c-c a^{2}+b a^{2}+b^{3}+b c^{2}-a b^{2}-b^{2} c-c a b+c a^{2} \\ &+c b^{2}+c^{3}-a b c-b c^{2}-c^{2} a \end{aligned}\end{gathered}

=

a

3

+ab

2

+ac

2

−ba

2

−abc−ca

2

+ba

2

+b

3

+bc

2

−ab

2

−b

2

c−cab+ca

2

+cb

2

+c

3

−abc−bc

2

−c

2

a

After simplification,

=a^{3}+b^{3}+c^{3}-3 a b c=a

3

+b

3

+c

3

−3abc

L.H.S=R.H.S

Hence, the given equation is proved, that the “left hand side” of the equation is equal to the “right hand side”.

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