Question

find the value of xy if x³+y³=12 and x+y=3​

Answer 1

Answer:

xy = $\dfrac{5}{3}$

Step-by-step explanation:

We have -

$x^{3}  + y^{3} = 12$

x + y = 3

xy = ?

We know that -

$(a+b)^{3}  = a^{3} + b^{3} + 3ab ( a+ b)$

So, $( x + y )^{3} = x^{3} + y^{3} + 3xy ( x + y )$

$3^{3} = 12 + 3xy \times 3$

27 = 12 + 9xy

27 - 12 = 9xy

9xy = 15

xy = $\dfrac{15}{9}$

xy = $\dfrac{5}{3}$

Answer 2

xy = \dfrac{5}{3}35

xy = \dfrac{5}{3}35Step-by-step explanation:

xy = \dfrac{5}{3}35Step-by-step explanation:We have -

xy = \dfrac{5}{3}35Step-by-step explanation:We have -x^{3}  + y^{3} = 12x3 +y3=12

xy = \dfrac{5}{3}35Step-by-step explanation:We have -x^{3}  + y^{3} = 12x3 +y3=12x + y = 3

xy = \dfrac{5}{3}35Step-by-step explanation:We have -x^{3}  + y^{3} = 12x3 +y3=12x + y = 3xy = ?

xy = \dfrac{5}{3}35Step-by-step explanation:We have -x^{3}  + y^{3} = 12x3 +y3=12x + y = 3xy = ?We know that -

xy = \dfrac{5}{3}35Step-by-step explanation:We have -x^{3}  + y^{3} = 12x3 +y3=12x + y = 3xy = ?We know that -(a+b)^{3}  = a^{3} + b^{3} + 3ab ( a+ b)(a+b)3 =a3+b3+3ab(a+b)

xy = \dfrac{5}{3}35Step-by-step explanation:We have -x^{3}  + y^{3} = 12x3 +y3=12x + y = 3xy = ?We know that -(a+b)^{3}  = a^{3} + b^{3} + 3ab ( a+ b)(a+b)3 =a3+b3+3ab(a+b)So, ( x + y )^{3} = x^{3} + y^{3} + 3xy ( x + y )(x+y)3=x3+y3+3xy(x+y)

xy = \dfrac{5}{3}35Step-by-step explanation:We have -x^{3}  + y^{3} = 12x3 +y3=12x + y = 3xy = ?We know that -(a+b)^{3}  = a^{3} + b^{3} + 3ab ( a+ b)(a+b)3 =a3+b3+3ab(a+b)So, ( x + y )^{3} = x^{3} + y^{3} + 3xy ( x + y )(x+y)3=x3+y3+3xy(x+y)3^{3} = 12 + 3xy \times 333=12+3xy×3

xy = \dfrac{5}{3}35Step-by-step explanation:We have -x^{3}  + y^{3} = 12x3 +y3=12x + y = 3xy = ?We know that -(a+b)^{3}  = a^{3} + b^{3} + 3ab ( a+ b)(a+b)3 =a3+b3+3ab(a+b)So, ( x + y )^{3} = x^{3} + y^{3} + 3xy ( x + y )(x+y)3=x3+y3+3xy(x+y)3^{3} = 12 + 3xy \times 333=12+3xy×327 = 12 + 9xy

xy = \dfrac{5}{3}35Step-by-step explanation:We have -x^{3}  + y^{3} = 12x3 +y3=12x + y = 3xy = ?We know that -(a+b)^{3}  = a^{3} + b^{3} + 3ab ( a+ b)(a+b)3 =a3+b3+3ab(a+b)So, ( x + y )^{3} = x^{3} + y^{3} + 3xy ( x + y )(x+y)3=x3+y3+3xy(x+y)3^{3} = 12 + 3xy \times 333=12+3xy×327 = 12 + 9xy27 - 12 = 9xy

xy = \dfrac{5}{3}35Step-by-step explanation:We have -x^{3}  + y^{3} = 12x3 +y3=12x + y = 3xy = ?We know that -(a+b)^{3}  = a^{3} + b^{3} + 3ab ( a+ b)(a+b)3 =a3+b3+3ab(a+b)So, ( x + y )^{3} = x^{3} + y^{3} + 3xy ( x + y )(x+y)3=x3+y3+3xy(x+y)3^{3} = 12 + 3xy \times 333=12+3xy×327 = 12 + 9xy27 - 12 = 9xy9xy = 15

xy = \dfrac{5}{3}35Step-by-step explanation:We have -x^{3}  + y^{3} = 12x3 +y3=12x + y = 3xy = ?We know that -(a+b)^{3}  = a^{3} + b^{3} + 3ab ( a+ b)(a+b)3 =a3+b3+3ab(a+b)So, ( x + y )^{3} = x^{3} + y^{3} + 3xy ( x + y )(x+y)3=x3+y3+3xy(x+y)3^{3} = 12 + 3xy \times 333=12+3xy×327 = 12 + 9xy27 - 12 = 9xy9xy = 15xy = \dfrac{15}{9}915

xy = \dfrac{5}{3}35Step-by-step explanation:We have -x^{3}  + y^{3} = 12x3 +y3=12x + y = 3xy = ?We know that -(a+b)^{3}  = a^{3} + b^{3} + 3ab ( a+ b)(a+b)3 =a3+b3+3ab(a+b)So, ( x + y )^{3} = x^{3} + y^{3} + 3xy ( x + y )(x+y)3=x3+y3+3xy(x+y)3^{3} = 12 + 3xy \times 333=12+3xy×327 = 12 + 9xy27 - 12 = 9xy9xy = 15xy = \dfrac{15}{9}915xy = \dfrac{5}{3}35

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