Question

If x and y are two positive real no such that 25x square 49ysquare equal841 and xy equal 12 then find the value of 125x cube 343ycube

Answer 1

25x^2 + 49y^2 = 841
and
125x^3 + 343y^3 can be factored as the "difference of cubes"
125x^3 + 343y^3 = (5x + 7y)(25x^2 - 35xy + 49y^2)
we know xy = 12
125x^3 + 343y^3 = (5x + 7y)(25x^2 - 35(12) + 49y^2)
125x^3 + 343y^3 = (5x + 7y)(25x^2 - 420 + 49y^2)
we know that 25x^2 + 49y^2 = 841
125x^3 + 343y^3 = (5x + 7y)(841 - 420)
125x^3 + 343y^3 = 421(5x +
if x and y are two positive real numbers such that 25x square + 49y s quare equal to 841 and xy equal to 12 then find the value of 125x cube + 343y cube
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841 = 400 + 441 = 25*4^2 + 49*3^2
----
x = 4, y = 3
---
125*4^3 + 343*3^3 = 8000 + 9261
= 17261

Answer 2

Answer:

125x^3 + 343y^3 = 421(5x + 7y)

Step-by-step explanation:

f x and y are two positive real numbers such that 25x square + 49y square equal to 841 and xy equal to 12 then find the value of 125x cube + 343y cube

:

25x^2 + 49y^2 = 841

and

125x^3 + 343y^3 can be factored as the "difference of cubes"

 

125x^3 + 343y^3 = (5x + 7y)(25x^2 - 35xy + 49y^2)

we know xy = 12

125x^3 + 343y^3 = (5x + 7y)(25x^2 - 35(12) + 49y^2)

125x^3 + 343y^3 = (5x + 7y)(25x^2 - 420 + 49y^2)

we know that 25x^2 + 49y^2 = 841

125x^3 + 343y^3 = (5x + 7y)(841 - 420)

125x^3 + 343y^3 = 421(5x + 7y)