If the sum of squares of two real numbers is | vedclass

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(C) Let the two real numbers be $x$ and $y$.Given that $x^{2} + y^{2} = 41$ and $x + y = 9$.We know that $(x + y)^{2} = x^{2} + y^{2} + 2xy$.Substitut

Vedclass · Accepted Answer

$189$

Vedclass · Answer

(C) Let the two real numbers be $x$ and $y$.Given that $x^{2} + y^{2} = 41$ and $x + y = 9$.We know that $(x + y)^{2} = x^{2} + y^{2} + 2xy$.Substituting the given values: $9^{2} = 41 + 2xy$.$81 = 41 + 2xy \implies 2xy = 40 \implies xy = 20$.We need to find the sum of cubes,which is $x^{3} + y^{3}$.Using the identity $x^{3} + y^{3} = (x + y)(x^{2} + y^{2} - xy)$.Substituting the values: $x^{3} + y^{3} = (9)(41 - 20)$.$x^{3} + y^{3} = 9 	imes 21 = 189$.

If the sum of squares of two real numbers is $41$ and their sum is $9,$ then the sum of cubes of these two numbers is

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