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:
Option 1: 169
Option 2: 209
Option 3: 189
Option 4: 198

Question

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:

Option 1: 169

Option 2: 209

Option 3: 189

Option 4: 198

Team Careers360 · Accepted Answer

Correct Answer: 189

Solution : Let the two numbers be $ ext{x}$ and $ ext{y}$.
As per the condition of the question,
⇒ $ ext{x}^{2} + ext{y}^{2}$ = 41 and $ ext{x + y} = 9$
We know $ ext{(x + y)}^{2} = ext{x}^{2} + ext{y}^{2} + ext{2xy}$
Putting values in the given expression:
$ ext{(9)}^{2} = 41 + ext{2xy}$
⇒ $ ext{2xy} = 81 - 41$
⇒ $ ext{xy} = \frac{40}{2} = 20$
Using the formula, $ ext{x}^{3} + ext{y}^{3} = ext{(x + y)}^{3}- ext{3xy} ext{(x+y)}$, we get,
$ ext{x}^{3} + ext{y}^{3} = ext{(9)}^{3}- 3 imes20 imes 9$
⇒ $ ext{x}^{3} + ext{y}^{3} = 729- 540 = 189$
Hence, the correct answer is 189.

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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:Option 1: 169Option 2: 209Option 3: 189Option 4: 198

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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:
Option 1: 169
Option 2: 209
Option 3: 189
Option 4: 198