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# Algebra

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172
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If x+y=9 and xy=18, what is the value of x^3+y^3?

Jun 1, 2022

#2
+2666
0

We have $$x^3 + y^3$$. Dividing this by $$(x+y)$$, we get $$(x^2 + y^2 - xy)$$,

Now, we recall the identity: $$x^2 + y^2 = (x+y)^2 - 2xy$$

Substituting this in gives us: $$(x+y)((x+y)^2 - 3xy)$$

Now, plugging in the values we know, we have: $$9 \times (81 - 3 \times 18) = \color{brown}\boxed{243}$$

Jun 2, 2022

#1
+14627
+1

What is the value of x^3+y^3?

Hello Guest!

$$x+y=9\\ xy=18\\ y=\frac{18}{x}=9-x\\ 18=9x-x^2\\ x^2-9x+18=0\\ x=4.5\pm \sqrt{4.5^2-18}=4.5\pm 1.5$$

$$x\in \{6,\ 3\} \\ y\in \{3,\ 6\}\\ \color{blue}x^3+y^3= 216+27=243$$

!

Jun 1, 2022
#2
+2666
0

We have $$x^3 + y^3$$. Dividing this by $$(x+y)$$, we get $$(x^2 + y^2 - xy)$$,

Now, we recall the identity: $$x^2 + y^2 = (x+y)^2 - 2xy$$

Substituting this in gives us: $$(x+y)((x+y)^2 - 3xy)$$

Now, plugging in the values we know, we have: $$9 \times (81 - 3 \times 18) = \color{brown}\boxed{243}$$

BuilderBoi Jun 2, 2022