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Question Number 32379 by mondodotto@gmail.com last updated on 24/Mar/18

Answered by MJS last updated on 24/Mar/18

$${(a-b)}^5=a^5-5a^{4}b+10a^3{b}^2-10a^2{b}^3+5{ab}^4-b^5$$$$a=\frac{x}{4y^3};b=\frac{2y}{x^2}$$$$a^5=\frac{x^5}{1024y^{15}}$$$$-5a^{4}b=-5\frac{x^4}{256y^{12}}\times \frac{2y}{x^2}=-\frac{5x^2}{128y^{11}}\Rightarrow$$$$\Rightarrow A=128;p=11$$$$10a^3{b}^2=10\frac{x^3}{64y^9}\times \frac{4y^2}{x^4}=\frac{5}{8{xy}^7}\Rightarrow$$$$\Rightarrow B=8$$

Commented by mondodotto@gmail.com last updated on 24/Mar/18

$$\mathrm{thanx}$$

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