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If the sum of square of two real numbers is 41, and their sum is 9. Then the sum of cubes of these two numbers is ?

If the sum of square of two real numbers is 41, and their sum is 9. Then the sum of cubes of these two numbers is ? Correct Answer 189

$$\eqalign{ & {\text{Let the numbers are a, b}} \cr & {a^2} + {b^2} = 41 \cr & a + b = 9 \cr & \Rightarrow {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \cr & \Rightarrow {9^2} = 41 + 2ab \cr & \Rightarrow 81 - 41 = 2ab \cr & \Rightarrow ab = 20 \cr & {\text{Take }} \cr & a = 5 \cr & b = 4 \cr & \Rightarrow {a^3} + {b^3} = {5^3} + {4^3} \cr & \Rightarrow {a^3} + {b^3} = 125 + 64 \cr & \Rightarrow {a^3} + {b^3} = 189 \cr} $$

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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
Count the number of cubes in the given figure.
Cubes and Dice mcq question image
Count the number of cubes in the given figure.
Cubes and Dice mcq question image