Question

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

A) 169 B) 209 C) 189 D) 198

Answer 1

Answer:-

$\sf{C) \ 189 \ is \ the \ correct \ option.}$

$\sf{The \ sum \ of \ the \ cubes \ of \ the \ number}$

$\sf{is \ 189.}$

Given:

• If the sum of squares of two real numbers is 41.
• And their sum is 9.

To find:

• The sum of cubes of these two numbers.

Solution:

$\sf{Let \ the \ numbers \ be \ x \ and \ y.}$

$\sf{According \ to \ the \ the \ first \ condition.}$

$\sf{\implies{x^{2}+y^{2}=41...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{\implies{x+y=9...(2)}}$

$\sf{According \ to \ the \ identity}$

$\boxed{\sf{(a+b)^{2}=a^{2}+b^{2}+2ab}}$

$\sf{(x+y)^{2}=x^{2}+y^{2}+2xy}$

$\sf{...from \ (1) \ and \ (2)}$

$\sf{\therefore{9^{2}=41+2xy}}$

$\sf{\therefore{2xy=81-41}}$

$\sf{\therefore{2xy=40}}$

$\sf{\therefore{xy=\frac{40}{2}}}$

$\sf{\therefore{xy=20...(3)}}$

$\sf{According \ to \ the \ identity.}$

$\boxed{\sf{a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)}}$

$\sf{\therefore{x^{3}+y^{3}=(x+y)^{3}-3xy(x+y)}}$

$\sf{...from \ (2) \ and \ (3)}$

$\sf{\therefore{x^{3}+y^{3}=9^{3}-3(20)(9)}}$

$\sf{\therefore{x^{3}+y^{3}=729-540}}$

$\sf{\therefore{x^{3}+y^{3}=189}}$

$\sf\purple{\tt{\therefore{The \ sum \ of \ the \ cubes \ of \ the \ number}}}$

$\sf\purple{\tt{is \ 189.}}$