Just for fun!-XIV

Algebra Level 1

Two reals x x and y y are such that x − y = 4 x-y=4 and x 3 − y 3 = 28 x^{3}-y^{3}=28 . Compute x y xy .

If the answer is − z -z , provide the answer as z z .


The answer is 3.

Swapnil Das by Swapnil Das , 20, India

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4 solutions

Department 8
Aug 21, 2015

We have ( x − y ) 3 = x 3 − y 3 − 3 x y ( x − y ) 64 = 28 − 3 x y ( 4 ) x y = − 3 (x-y)^{3}=x^ {3}-y^{3}-3xy(x-y) \\ 64=28-3xy(4) \\ xy=-3\\ Therefore answer is 3 \boxed{3}

10 Helpful 0 Interesting 0 Brilliant 0 Confused

28 = ( x − y ) ( x 2 + x y + y 2 ) = 4 ∗ { ( x − y ) 2 + 3 x y } ∴ 7 = 16 + 3 x y ⟹ − 9 = 3 x y . x y = − 3 28=(x-y)(x^2+xy+y^2)=4*\{(x-y)^2+3xy\}\\ \therefore~7=16+3xy\\ \implies~-9=3xy.\\ xy=~-~3

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Chandresh Shah
Aug 23, 2015

X cube-Y cube =28, [X-Y] cube = X cube - Y cube-3xy[x-y].Putting the value x-y =4 in above equation. cube of 4=28-3xy[4]. 64=-12xy+28. 64-28=12xy 36=-12xy,
xy=36/-12=-3

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Caeo Tan
Aug 22, 2015

from x-y=4, we get (x-y)^3=64

x^3-3(x^2)y+3x(y^2)-y^3=64

28-3xy(x-y)=64

28-12xy=64

-12xy=36

xy=-3

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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