Question

(x+y)=5 and xy=4 then x3+y3 is equal to

Answer 1

Solution:-

Given :-

=> ( x + y ) = 5

=> xy = 4

To find

=> x³ + y³

Using this identity

( x + y )³ = x³ + y³ + 3xy( xy )

Now put the value

( x + y )³ = x³ + y³ + 3xy( x + y )

(5)³ = x³ + y³ + 3 × 4 ( 5 )

125 = x³ + y³ + 3 × 20

125 = x³ + y³ + 60

x³ + y³ = 125 - 60

x³ + y³ = 65

Answer:- 65

Some identities

=> ( a + b )² = a² + b² + 2ab

=> ( a - b )² = a² + b² - 2ab

=> ( a² - b² ) = ( a - b )(a + b )

=> ( a + b )³ = a³ + b³ + 3ab( a + b )

=> ( a - b )³ = a³ - b³ - 3ab( a - b )

Answer 2

Answer:

$\huge \sf { \underline{given \ \: : }}$

$\bull \bf(x + y) = 5$

$\bull \bf( xy )= 4$

$\huge \sf { \underline{ to \: find\ \: : }}$

$\bf \bull the \: value \: of \: {x}^{3} \: and \: {y}^{3}$

$\huge \sf { \underline{let \: us \: solve \ \: : }}$

$\bf {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)$

$\bf \to {5}^{3} = {x}^{3} + {y}^{3} + 3 \times (4 \times 5)$

$\bf \to125 = {x}^{3} + {y}^{3} + 3 \times (20)$

$\bf \to 125 = { x }^{3} + {y}^{3} + 60$

$\bf \to {x}^{3} + {y}^{3} = 125 - 60$

$\boxed{\red{\sf{{ {{x}^{3} + {y}^{3} = 65 }}}}}$

$\huge \sf { \underline{Identity\:used\ \: : }}$

$\bf {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)$