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KingAce

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PROBLEM: if the sum of two numbers if 8, and the sum of their cubes is 20, what is the sum of their sqaures?

- is there any way to get the two numbers besides simply guessing-and-checking?
Also, could I combine the equations: x+y=8 and x^3+y^3=20 or manipulate them to arrive at the answer? thanks :)
 
This is a thinky question - NOT a "trick" question. It is designed to make you think about things you should have learned.

Given
x^3 + y^3 = 20
x + y = 8

A little factoring and rearranging
x^3 + y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x^2 + y^2) - xy)
Substituting
(x^3 + y^3) = (x+y)((x^2 + y^2) - xy)
20 = (8)((x^2 + y^2) - xy) -- Wow!! If we knew xy, we would be done!

Something Else to Think About
(x+y)^3 = 8^3, but
(x+y)^3 = x^3 + 3*x^2*y + 3*x*y^2 + y^3 = (x^3 + y^3) + 3*x*y*(x+y)

More substituting
(x+y)^3 = (x^3 + y^3) + 3*x*y*(x+y)
8^3 = (20) + 3*x*y*(8) -- Well, perhaps we DO know the value of xy.

Resist the temptation to solve for 'x' and 'y'. You will not like the values you find.
 
x+y=8
x^3+y^3=20 factor

[x+y] [ x^2-xy+y^2=20 but x+y=8
x^2-xy+y^2= 20/8

square x+y
x^2+2xy+y^2=64
multiply the factored equation by 2
2x^2-2xy+2y^2=5
add the equations
3x^2+3y^2=69
x^2+y^2=23 answer

Arthur
 
I love it when different methodologies produce the same, unique answer. This should demonstrate to the student that it is important to STOP WORRYING about the right way to go about it. Just think about it and try something. Some ways are easier than others but this does NOT make one way wrong or inappropriate.
 
Hello, KingAce!

There is a sneaky solution to this problem.
. . Hmmm, tkhunny beat me to it.

Of the sum of two numbers if 8, and the sum of their cubes is 20,
what is the sum of their sqaures?
Click to expand...

We are given:\(\displaystyle \;\begin{array}{c}(1)\\(2)\end{array}\;\begin{array}{ccc}\;x\,+\,y & = & 8 \\ (2)\;x^3\,+\,y^3 & = & 20\end{array}\)


Cube (1): . . . . . \(\displaystyle (x\,+\,y)^3 \:= \:8^3\)

. . \(\displaystyle x^3\,+\,3x^2y\,+\,3xy^2\,+\,y^3\:=\:512\)

. . \(\displaystyle x^3\,+\,y^3\,+\,3x^2y\,+\,3xy^2 \:=\:512\)

. . \(\displaystyle \underbrace{(x^3\,+\,y^3)}_{\downarrow}\,+\,3xy\underbrace{(x\,+\,y)}_{\downarrow}\:=\:512\)
. . . . . \(\displaystyle 20\;\;\;\;+\;\;\;3xy(8) \;\;=\;512\)

. . which simplifies to: \(\displaystyle \,2xy\,=\,41\;\) (3)


Square (1): . . .\(\displaystyle (x\,+\,y)^2\:=\:8^2\)

. . . . . . . \(\displaystyle x^2\,+\,\underbrace{2xy}_{\downarrow}\,+\,y^2\:=\:64\)
From (3): \(\displaystyle \;x^2\,+\,41\,+\,y^2\:=\:64\)


Therefore: \(\displaystyle \:x^2\,+\,y^2\:=\:23\)

 
See how it goes? Arthur and Soroban picked a slightly simpler methodology than the one that came first to my mind. Unique is unique!
 
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