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If a, b, c and d are in proportion, then show that `(a^(3)+3ab^(2))/(3a^(2)b+b^(3))=(c^(3)+3cd^(2))/(3c^(2)d+d^(3))`

Answer» Given a, b, c and d are in proportion
` rArr a/b = c/d `
Applying componendo and dividendo,
` (a+b)/(a-b)= (c+d)/(c-d)`
Cubing on both sides, we have
`((a+b)^(3))/((a-b)^(3)) = [(c+d)/(c-d)]^(3)`
` rArr (a^(3)+3a^(2)b+3ab^(2)+b^(3))/(a^(3)-3a^(3)b+3ab^(2)-b^(3))=(c^(3)+3c^(2)d+3cd^(2)+l^(3))/(c^(3)-3c^(2)d+3cd^(2)-d^(3))`
`((a^(3)+3ab^(2))+(3a^(2)b+b^(3)))/((a^(3)+3ab^(2))-(3a^(2)b+b^(3)))=((c^(3)+3cd^(2))+(3c^(2)d+d^(3)))/((c^(3)+3cd^(2))-(3c^(2)d+d^(3)))`
` rArr (a^(3)+3ab^(2))/(3a^(2)b+b^(3))=(c^(3)+3cd^(2))/(3c^(2)d+d^(3))`
Hence, proved.


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