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
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Correct Answer: 189
Solution : Let the two numbers be $\text{x}$ and $\text{y}$. As per the condition of the question, ⇒ $\text{x}^{2} + \text{y}^{2}$ = 41 and $\text{x + y} = 9$ We know $\text{(x + y)}^{2} = \text{x}^{2} + \text{y}^{2} + \text{2xy}$ Putting values in the given expression: $\text{(9)}^{2} = 41 + \text{2xy}$ ⇒ $\text{2xy} = 81 - 41$ ⇒ $\text{xy} = \frac{40}{2} = 20$ Using the formula, $\text{x}^{3} + \text{y}^{3} = \text{(x + y)}^{3}- \text{3xy}\text{(x+y)}$, we get, $\text{x}^{3} + \text{y}^{3} = \text{(9)}^{3}- 3\times20\times 9$ ⇒ $\text{x}^{3} + \text{y}^{3} = 729- 540 = 189$ Hence, the correct answer is 189.
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