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Mathematics

If x3+3xy23x2y+y3=m3+3mn23m2n+n3\dfrac{x^3 + 3xy^2}{3x^2y + y^3} = \dfrac{m^3 + 3mn^2}{3m^2n + n^3},

show that : nx = my.

Ratio Proportion

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Answer

Given,

x3+3xy23x2y+y3=m3+3mn23m2n+n3\dfrac{x^3 + 3xy^2}{3x^2y + y^3} = \dfrac{m^3 + 3mn^2}{3m^2n + n^3}

Applying componendo and dividendo,

⇒x3+3xy2+3x2y+y3x3+3xy2−(3x2y+y3)=m3+3mn2+3m2n+n3m3+3mn2−(3m2n+n3)⇒(x+y)3(x−y)3=(m+n)3(m−n)3⇒(x+y)(x−y)=(m+n)(m−n)\Rightarrow \dfrac{x^3 + 3xy^2 + 3x^2y + y^3}{x^3 + 3xy^2 - (3x^2y + y^3)} = \dfrac{m^3 + 3mn^2 + 3m^2n + n^3}{m^3 + 3mn^2 - (3m^2n + n^3)} \\[1em] \Rightarrow \dfrac{(x + y)^3}{(x - y)^3} = \dfrac{(m + n)^3}{(m - n)^3} \\[1em] \Rightarrow \dfrac{(x + y)}{(x - y)} = \dfrac{(m + n)}{(m - n)}

Applying componendo and dividendo,

⇒x+y+x−yx+y−(x−y)=m+n+m−nm+n−(m−n)⇒2x2y=2m2n⇒xy=mn⇒nx=my.\Rightarrow \dfrac{x + y + x - y}{x + y - (x - y)} = \dfrac{m + n + m - n}{m + n - (m - n)} \\[1em] \Rightarrow \dfrac{2x}{2y} = \dfrac{2m}{2n} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{m}{n} \\[1em] \Rightarrow nx = my.

Hence, proved that nx = my.

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