Question

ii) If sum of two numbers is 6, and the sum of
their cubes is 36, find the sum of their squares​

Answer 1

Answer:

this will be the answer for your question

Answer 2

let the 2 no.s be x and y

$x + y = 6 \\ {x}^{2} + {y}^{2} + 2xy = 36 \\ {x}^{2} + {y}^{2} = 36 - 2xy .............(1)\\ and \\ {x}^{3} + {y}^{3} = 36 \\ (x + y)( {x}^{2} - xy + {y}^{2} ) = 36 \\ 6({x}^{2} - xy + {y}^{2} ) = 36 \\ {x}^{2} - xy + {y}^{2} = 6 \\ 36 - 2xy - xy = 6 \\ 30 = 3xy \\ xy = 10 \\ \\ putting \: xy \: in \: eq1 \\ {x}^{2} + {y}^{2} = 36 - 20 \\ {x}^{2} + {y}^{2} = 16$