Question

if x2+y2 =29 and xy=10.then find the value of x3-y3 ,when x is greater than y

Answer 1

x^2+y^2=29 and xy=10
so..
${x}^{2} + {y}^{2} = 29 \\ or \: \: {(x - y)}^{2} + 2xy = 29 \\ or \: \: {(x - y)}^{2} + 2 \times 10 = 29 \\ or \: \: {(x - y)}^{2} = 9 \\ or \: \: (x - y) = 3$
so x-y = 3
hence the value of x^3 - y^3 will be ----
${x}^{3} - {y}^{3} \\ = {(x - y)}^{3} + 3xy(x - y) \\ = {3}^{3} + 3 \times 10 \times 3 \\ = 27 + 90 \\ = 117(answer)$
hope it helps, please mark it brainliest answer...

Answer 2

Answer:

57

Step-by-step explanation:

x²+y²=29 (given),

xy=10 (given),

So, x–y=3

By the equation,

x³–y³=(x–y)(x²–xy+y²)

=3×19

=57