Question
Download Solution PDFIf x2 + y2 + z2 = xy + yz + zx and x = 1, then find the value of \(\rm \frac{10x^4+5y^4+7z^4}{13x^2y^2+6y^2z^2+3z^2x^2}\)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
x = 1
x2 + y2 + z2 = xy + yz + zx
Calculations:
x2 + y2 + z2 - xy - yz - zx = 0
⇒(1/2)[(x - y)2 + (y - z)2 + (z - x)2] = 0
⇒x = y , y = z and z = x
But x = y = z = 1
so, \(\rm \frac{10x^4+5y^4+7z^4}{13x^2y^2+6y^2z^2+3z^2x^2}\)
= {10(1)4 + 5(1)4 + 7(1)4}/{13(1)2(1)2+ 6(1)2(1)2 + 3(1)2(1)2}
= 22/22
= 1
Hence, the required value is 1.
Last updated on Jun 11, 2025
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