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Question Number 51220 by Gulay last updated on 25/Dec/18

Commented by Gulay last updated on 25/Dec/18

sir help me pls

$$\mathrm{sir}\:\mathrm{help}\:\mathrm{me}\:\mathrm{pls} \\ $$$$ \\ $$

Answered by afachri last updated on 25/Dec/18

55 = x 52 = y (x − y )^3 = x^2 − 3x^2 y + 3xy^2 − y^3 (55−52)^3 = 55^3 − 3(55^2 )(52) + 3(55)(52^2 )− 52^3 the product in RHS with evaluate LHS. (55 − 52)^3 = 3^3 = 27

$$\mathrm{55}\:=\:{x} \\ $$$$\mathrm{52}\:=\:{y} \\ $$$$\left({x}\:−\:{y}\:\right)^{\mathrm{3}} \:=\:{x}^{\mathrm{2}} −\:\mathrm{3}{x}^{\mathrm{2}} {y}\:+\:\mathrm{3}{xy}^{\mathrm{2}} −\:{y}^{\mathrm{3}} \\ $$$$\left(\mathrm{55}−\mathrm{52}\right)^{\mathrm{3}} =\:\mathrm{55}^{\mathrm{3}} −\:\mathrm{3}\left(\mathrm{55}^{\mathrm{2}} \right)\left(\mathrm{52}\right)\:+\:\mathrm{3}\left(\mathrm{55}\right)\left(\mathrm{52}^{\mathrm{2}} \right)−\:\mathrm{52}^{\mathrm{3}} \\ $$$$\mathrm{the}\:\mathrm{product}\:\mathrm{in}\:\mathrm{RHS}\:\mathrm{with}\:\mathrm{evaluate}\:\mathrm{LHS}. \\ $$$$\left(\mathrm{55}\:−\:\mathrm{52}\right)^{\mathrm{3}} =\:\mathrm{3}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{27}\: \\ $$

Commented by Gulay last updated on 25/Dec/18

thank you so much sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$$$ \\ $$

Commented by afachri last updated on 25/Dec/18

ur welcome, Sir

$$\mathrm{ur}\:\mathrm{welcome},\:\mathrm{Sir} \\ $$

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