Math, asked by mmcian, 1 year ago

if x^2+y^2=49 and x-y=3.find the value of x^3-y^3

Answers

Answered by Anonymous
15
x^3 - y^3 = (x-y) ( x^2 + y^2 +xy)

we need xy


(x-y)^2 = x^2 + y^2 - 2xy

9 = 49 - 2xy

xy = 20

x^3 - y^3 = (x-y) ( x^2 + y^2 +xy) = 3 ( 49 - 20) = 2 (29) = 87

Anonymous: mark as brainliest
Answered by Anonymous
25
given \\ {x }^{2} + {y}^{2} = 49 \\ and \: x - y = 3 \\ we \: have \: to \: find \: \: {x}^{3} - {y}^{3} \\ now \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ putting \: value \: of \: x - y \: and \: {x}^{2} + {y}^{2}\\ from \: the \: question \\ we \: get \\ {3}^{2} = 49 - 2xy \\ 2xy = 40 \\ xy = 20 - - - - - 1 \\ now \: using \: identity \\ {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + {b}^{2} + ab) \\ hence \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy) \\ putting \: all \: values \: we \: get \\ {x}^{ 3 } - {y}^{3} = 3 \times (49 + 20) \\ 3 \times 69 = 207 \\ \\ hope \: this \: helps \: you
Similar questions