if x2+y2=29 and xy=10, then find the value of x3-y3

x2+y2 = 29x-y2+2xy = 29x-y2+2×10 = 29x-y2 = 29-20x-y2 = 9x-y= 9 = 3         ...(i)x-y3 = 33     {Cubing both sides}x3-y3-3xyx-y = 27    {using the identity a-b3 = a3-b3-3aba-b}x3-y3-3×10×3 = 27      {using (i) and given xy = 10}x3-y3-90 = 27x3-y3 = 117

  • 17

(x-y)2=x2+y2-2xy=49

x-y=7

(x-y)3=x3-y3-3xy(x-y)

343=x3-y3-210

x3-y3=553

  • -3

you are wrong . we have to use identites.

  • 7
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