·
Verify:
a'+b+c-3abc =
(a+b+c) [(a-b) + (b-c)' +(c-a)
Answers
Answer:
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”.