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Question
If `(x^3 + 3xy^2)/(3x^2y + y^3) = (m^3 + 3mn^2)/(3m^2n + n^3)`, show that nx = my.
Solution
`(x^3 + 3xy^2)/(3x^2y + y^3) = (m^3 + 3mn^2)/(3m^2n + n^3)`
Applying componendo and dividendo,
`(x^3 + 3xy^2 + 3x^2y + y^3)/(x^3 + 3xy^2 - 3x^2y - y^3) = (m^3 + 3mn^2 + 3m^2n + n^3)/(m^3 + 3mn^2 - 3m^2n - n^3)`
`(x + y)^3/(x - y)^3 = (m + n)^3/(m - n)^3`
`(x + y)/(x - y) = (m + n)/(m - n)`
Applying componendo and dividendo,
`(x + y + x - y)/(x + y - x + y) = (m + n + m - n)/(m + n - m + n)`
`(2x)/(2y) = (2m)/(2n)`
`x/y = m/n`
nx = my
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