Question

if x = 1001, y = 1002 and 2 = 1003, find the value of x + y + z° - 3xyz, without
actually calculating cubes.
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Answer 1

Answer:

x=1001=1002–1=y-1

z=1003=1002+1=y+1

Now x^3+y^3+z^3–3x.y.z=?

put x= y-1 and z= y+1

= (y-1)^3+y^3+(y+1)^3–3(y-1).y.(y+1)

= y^3–3y(y-1)-1+y^3+y^3+3y(y+1)+1–3y(y^2–1)

= 3y^3–3y(y-1-y-1+y2–1)

= 3y^3–3y(y^2–3)

= 3y^3–3y^3+9y

= 9y

put y = 1002

= 9×1002

= 9018