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ICSE Class 10 Mathematics Question 18 of 21

Solved 2025 Specimen Paper ICSE Class 10 Mathematics — Question 13

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13
Question

Question 8(ii)

If x, y and z are in continued proportion, prove that :

xy2.z2+yz2.x2+zx2.y2=1x3+1y3+1z3\dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} = \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3}

Answer

Given,

x, y and z are in continued proportion.

∴xy=yz⇒y2=xz\therefore \dfrac{x}{y} = \dfrac{y}{z} \\[1em] \Rightarrow y^2 = xz

To prove :

xy2.z2+yz2.x2+zx2.y2=1x3+1y3+1z3\dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} = \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3}

Solving L.H.S.,

⇒xy2.z2+yz2.x2+zx2.y2⇒x3+y3+z3x2.y2.z2⇒x3+y3+z3x2.xz.z2⇒x3+y3+z3x3.z3⇒x3x3.z3+y3x3z3+z3x3.z3⇒1z3+y3(xz)3+1x3⇒1z3+y3(y2)3+1x3⇒1z3+y3y6+1x3⇒1z3+1y3+1x3.\Rightarrow \dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} \\[1em] \Rightarrow \dfrac{x^3 + y^3 + z^3}{x^2.y^2.z^2} \\[1em] \Rightarrow \dfrac{x^3 + y^3 + z^3}{x^2.xz.z^2} \\[1em] \Rightarrow \dfrac{x^3 + y^3 + z^3}{x^3.z^3} \\[1em] \Rightarrow \dfrac{x^3}{x^3.z^3} + \dfrac{y^3}{x^3z^3} + \dfrac{z^3}{x^3.z^3} \\[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{y^3}{(xz)^3} + \dfrac{1}{x^3} \\[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{y^3}{(y^2)^3} + \dfrac{1}{x^3} \\[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{y^3}{y^6} + \dfrac{1}{x^3} \\[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{1}{y^3} + \dfrac{1}{x^3}.

Since, L.H.S. = R.H.S.

Hence, proved that xy2.z2+yz2.x2+zx2.y2=1x3+1y3+1z3\dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} = \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3}.