Question

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Answer 1

1.

$\sf Let~ \frac{3}{x} ~be~m~and~\frac{5}{y}~be n \\\\\sf We \: can \: write \: it \: as \: \\\\\sf (m+n) (n^2+m^2-mn) ~which \: equals \: m^3-n^3 \\\\\sf => \big(\frac{3}{x} \big) ^3-\big(\frac{5}{y} \big)^3 \\\\\sf \big(\frac{27}{x^3} \big)-\big(\frac{125}{y^3} \big)\\\\\sf but~ given ~answer~ is~ \big(\frac{9a}{x^3} \big)+\big(\frac{5b}{y^3} \big)$

Comparing both of them we get:

$9a=27 \\\\\sf a= \frac{27}{9} = 3$

$5b= -125 \\\\\sf b= -25$

We have to find:

$\frac{b+a-1}{3} \\\\\sf \frac{-25+3-1}{3}=>\frac{-23}{3}$