Question

Using suitable identify, prove that
(0.52)^3+(0.62)^3-(0.14)^3+3(0.52)(0.62)(0.14)/(0.52)^2+(0.62)^2+(0.14)^2-(0.52)(0.62)+(0.62)(0.14)+(0.52)(0.14)=1​

Answer 1

Step-by-step explanation:

Numerator = (.52)3 + (.62)3 -(.14)3 - 3(.52)(.62) -(-.14)

this is of the form

x3 + y3 +z3 - 3xyz where x = .52, y =.62 z =-.14

x3 + y3 +z3 - 3xyz =(x+y+z)(x2 +y2 +z2 -xy -yz -zx)

(.52)3 + (.62)3 -(.14)3 - 3(.52)(.62) -(-.14) = ( .52 + .62 +(-.14)) ( (.52)2 + (.62)2 + (-.14)2 -(.52)(.62) -(.62)(-.14) -(-.14)(.52) )

= (1.14 - .14)(( (.52)2 + (.62)2 + (-.14)2 -(.52)(.62) +(.62)(.14) + (.14)(.52) )

=(1) ( (.52)2 + (.62)2 + (-.14)2 -(.52)(.62) +(.62)(.14) + (.14)(.52) )

=( (.52)2 + (.62)2 + (-.14)2 -(.52)(.62) +(.62)(.14) + (.14)(.52) )

This is the denominator

thus {(.52)3 + (.62)3 -(.14)3 - 3(.52)(.62) -(-.14)}/( (.52)2 + (.62)2 + (-.14)2 -(.52)(.62) +(.62)(.14) + (.14)(.52) ) = 1