256
Question
Prove the following identities:
4∑(b−c)(b+c−2a)3=9∑(b−c)(b+c−a)2.
4∑(b−c)(b+c−2a)3=9∑(b−c)(b+c−a)2.
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Solution
⇒(b−c)(b+c−2a)2+(c−a)(c+a−2b)2+(a−b)(a+b−2c)2=−9(b−c)(c−a)(a−b)
and (b−c)(b+c−a)2+(c−a)(c+a−b)2+(a−b)(a+b−c)2=−4(b−c)(c−a)(a−b)and, hence the above relation follows,4∑(b−c)(b+c−2a)2=9∑(b−c)(b+c−a)2
⇒(b−c)(b+c−2a)2+(c−a)(c+a−2b)2+(a−b)(a+b−2c)2
=−9(b−c)(c−a)(a−b)
and (b−c)(b+c−a)2+(c−a)(c+a−b)2+(a−b)(a+b−c)2
=−4(b−c)(c−a)(a−b)
and, hence the above relation follows,
4∑(b−c)(b+c−2a)2
=9∑(b−c)(b+c−a)2
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