Question

If x/a = y/b = z/c, prove that x^3/a^2 + y^3/b^2 + z^3/c^2 = (x+y+z)^3/ (a+b+c)^2

Answer 1

Step-by-step explanation:

$let \: \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k \\ x = ak \\ y = bk \\ z = ck \\ now \\ lhs = \frac{ {x}^{3} }{ {a}^{2} } + \frac{ {y}^{3} }{ {b}^{2} } + \frac{ {z}^{3} }{ {c}^{2} } \\ = \frac{ {a}^{3} \times {k}^{3} }{ {a}^{2} } + \frac{ {b}^{3} \times {k}^{3} }{ {b}^{2} } + \frac{ {c}^{3} \times {k}^{3} }{ {c}^{2} } \\ = {k}^{3} (a + b + c) \\ \\ rhs = \frac{( {x + y + z)}^{3} }{ {(a + b + c)}^{2} } \\ = \frac{( {ak + bk + ck)}^{3} }{ {(a + b + c)}^{2} } \\ = \frac{ {k}^{3} \times ( {a + b + c)}^{3} }{( {a + b + c)}^{2} } \\ = {k}^{3} \times (a + b + c) \\ \\ \\ hence \: lhs = rhs \: \: \: \:( proved)$

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