Question

if x-y=4 and xy= 7 then find the value of x³-y³​

Answer 1

Answer:

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Answer 2

Answer:

This can be solved as a system of equations. First, solve for one of the variables in the first equation; we’ll choose to solve for x:

x + y = 7

Subtract y from both sides:

x = 7-y

Now substitute for x in the second equation:

(x^3) + (y^3) + 21xy

(7-y)^3 + (y^3) + 21*(7-y)y

First, expand the parentheses on the left:

((7-y)(7-y)(7-y)) + (y^3) + 21(7-y)y

Now use the FOIL method on the three sets of parentheses on the left; do the first two binomials first. You end up with:

(49 - 14y + y^2)(7-y) + (y^3) + 21(7-y)y

Now, distribute the binomial to the trinomial (both, again, are on the left; I’ve re-written the terms slightly to follow the standard format):

(7-y)(y^2 - 14y + 49) + (y^3) + 21(7-y)y

7y^2 - 98y + 343 - (y^3) + 14y^2 - 49y + (y^3) + 21(7-y)y

Now, combine like terms and re-write the equation to follow the standard format (ax + b + c):

-(y^3) + 21y^2 - 147y + 343 + (y^3) + 21(7-y)y

Note we have two instances of the term y^3; combine them; (y^3) and -(y^3). You end up with zero:

21(y^2) - 147y + 343 + 21y(7-y)

Now, use the distributive property on the binomial on the far right of the equation:

21(y^2) - 147y + 343 + 147y - 21(y^2)

Notice we have two instances of the same term, one positive and one negative: 21(y^2) and -21(y^2). These, again, combine to make zero. You are left with:

-147y + 147y + 343

Once again, two terms eliminate. You are left with 343.

That’s it!

I often make small errors in my answers to math questions, so if you see one please do kindly point it out in the comments and I’ll edit accordingly.

I hope this works for you! :)