Question

if the sum and the product of the two numbers are 7 and 45÷7 respectively find the sum of their cubes

Answer 1

let the numbers be x and y
$x + y = 7 - - - (1) \\ xy = \frac{45}{7} - - - (2) \\ from \: eq.1 \\ x = 7 - y \\ put \: this \: in \: eq.2 \\ (7 - y)(y) = \frac{45}{7} \\ 7y - {y}^{2} = \frac{45}{7 } \\ - 7{y}^{2} + 49y - 45 = 0 \\ 7 {y}^{2} - 49y + 45 = 0 \\ y = \frac{49 + \sqrt{ {49}^{2} - 4(7)(45) } }{14} \\ x = \frac{49 - \sqrt{ {(49)}^{2} - 4(7)(45)} }{14}$

Answer 2

Answer:

427/4

Step-by-step explanation:

here, a+b=7, ab=45/4 find a^3+b^3