If a + b + c = 0 show a^2 + b^2 + c^2 = 3abc
Answers
Step-by-step explanation:
Let us consider a quadratic polynomial p x^{2}+q x+r=0px
2
+qx+r=0
We know that if discriminant q^{2}-4 p r=0q
2
−4pr=0 , the roots of the equation are equal.
Here q=-2\left(a^{2}-b c\right)q=−2(a
2
−bc) ; p=c^{2}-a bp=c
2
−ab ; r=b^{2}-a cr=b
2
−ac
Therefore, discriminant=0
\begin{gathered}\begin{array}{l}{q^{2}=4 p r} \\ \\{\left(-2\left(a^{2}-b c\right)\right)^{2}=4\left(c^{2}-a b\right)\left(b^{2}-a c\right)} \\ \\{4\left(a^{4}+b^{2} c^{2}-2 a^{2} b c\right)=4\left(b^{2} c^{2}+a^{2} b c-a b^{3}-a c^{3}\right)} \\ \\{a^{4}+b^{2} c^{2}-2 a^{2} b c=b^{2} c^{2}+a^{2} b c-a b^{3}-a c^{3}}\end{array}\end{gathered}
q
2
=4pr
(−2(a
2
−bc))
2
=4(c
2
−ab)(b
2
−ac)
4(a
4
+b
2
c
2
−2a
2
bc)=4(b
2
c
2
+a
2
bc−ab
3
−ac
3
)
a
4
+b
2
c
2
−2a
2
bc=b
2
c
2
+a
2
bc−ab
3
−ac
3
\begin{gathered}\begin{array}{l}{a^{4}+b^{2} c^{2}-2 a^{2} b c-b^{2} c^{2}-a^{2} b c+a b^{3}+a c^{3}=0} \\ \\{a\left(a^{3}+b^{3}+c^{3}\right)=3 a^{2} b c} \\ \\{a\left(a^{3}+b^{3}+c^{3}-3 a b c\right)=0}\end{array}\end{gathered}
a
4
+b
2
c
2
−2a
2
bc−b
2
c
2
−a
2
bc+ab
3
+ac
3
=0
a(a
3
+b
3
+c
3
)=3a
2
bc
a(a
3
+b
3
+c
3
−3abc)=0
\begin{gathered}\begin{array}{l}{a=0 \text { or } a^{3}+b^{3}+c^{3}-3 a b c=0} \\ \\{a=0 \text { or } a^{3}+b^{3}+c^{3}=3 a b c}\end{array}\end{gathered}
a=0 or a
3
+b
3
+c
3
−3abc=0
a=0 or a
3
+b
3
+c
3
=3abc
Answer:
Question is wrong
I will tell you how it came.
Step-by-step explanation:
Cubing on both sides.
a+b = -c
Hence proved.
If you like the process please comment.