Question

The sum of two numbers is 7 and the sum of their cubes is 133 find the sum of their squares.

Answer 1

Let one number is x and other is (7-x)


$x {}^{3} + (7 - x {}^{} ) {}^{3} = 133$

$(x + 7 - x)(x {}^{2} + (7 - x) {}^{2} - x(7 - x)) = 133$
$7(x {}^{2} + 49 + x {}^{2} - 14x - 7x + x {}^{2} ) = 133$

$7(3x {}^{2} - 21x + 49) = 133$


$21x {}^{2} - 147x + 343 = 133$

$21x {}^{2} - 147x + 210 = 0$

Take 21 common...

$x {}^{2} - 7x + 10 = 0$

$x {}^{2} - 5x - 2x + 10 = 0$

$x(x - 5) - 2(x - 5) = 0$

$(x - 2)(x - 5) = 0$

$x = 2 \: \: and \: 5$

Hence no. are 2 and 5

Sum of their squares are

$2 {}^{2} + 5 {}^{2}$

$4 + 25 = 29$

$no \: is \: 29$


Thanks

Answer 2

$x + y = 7 \\ {x}^{3} + {y}^{3} = 133$
Since the sum of the numbers is quite small, you can use the hit and trial method to get the answers as 2 and 5. But for subjective purposes, here is the solution
${x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy) \\133 = 7( {x}^{2} + {y}^{2} - xy) \\ 19 = {x}^{2} + {y}^{2} - xy \\ {x}^{2} + {y}^{2} = 19 + xy$
Now consider this
$x + y = 7 \\ {x}^{2} + {y}^{2} + 2xy = 49 \\ {x}^{2} + {y}^{2} = 49 - 2xy \\ 49 - 2xy = 19 + xy \\ 3xy = 30 \\ xy = 10$
Put this is any of the two equations to get
${x}^{2} + {y}^{2} = 19 + xy \\ {x}^{2} + {y}^{2} = 19 + 10 \\ {x}^{2} + {y}^{2} = 29$
Or you can see that 10 can only be factored as 5 × 2 and therefore the only possible answers are 5 and 2!!!

Please give brainliest answer as I want to level up!!!