Question

If x = 19
y= 18

Then find value of :-
$\frac{x {}^{2} + y {}^{2} + xy }{x {}^{3 } - y {}^{3} }$
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Answer 1

Answer:

$\huge\underline{ \underline{ \red{your \: \: answer}}}$

Step-by-step explanation:

$\sf\red{ {a}^{3} - {b}^{3} = (a - b) ( {a}^{2} + ab + {b}^{2} )}\\ \\ \sf\pink{here \: x = 19 \: y = 18} \\ \\ \sf\red{ \frac{ {x}^{2} + {y}^{2} + xy }{ {x}^{3} - {y}^{3} } } \\ \\ \sf\green{ = > \frac{ {x}^{2} + xy + {y}^{2} }{(x - y)( {x}^{2} + xy + {y}^{2} )} } \\ \\ \sf\ \orange{ = > \frac{1}{(x - y)} } \\ \\ \sf\blue{ = > \frac{1}{19 - 18} } \\ \\ \sf\red{ = > \frac{1}{1} } \\ \\ \sf\purple{ = > 1} \\ \\ \\ \large\boxed{ \fcolorbox{green}{yellow}{hope \: it \: help \: you} }$

Answer 2

Step-by-step explanation:

$given \: \\ x = 19 \\ y = 18 \\ Then find value of :-</p><p>\frac{x {}^{2} + y {}^{2} + xy }{x {}^{3 } - y {}^{3} } \\ = > \frac{x {}^{2} + y {}^{2} + xy \: }{(x - y)(x {}^{2} + y {}^{2} + xy)} \\ = > \frac{1}{x - y} \\ = > \frac{1}{19 - 18} \\ \frac{1}{1} = 1$