Express ##x^3+y^3## in terms of ##m## and ##n##

[brotherbobby]

Homework Statement

If ##x+y=m## and ##x-y=n##, express ##\boldsymbol{x^3+y^3}## in terms of ##\boldsymbol m## and ##\boldsymbol n##.

Relevant Equations

We have ##a^3+b^3=(a+b)(a^2-ab+b^2)##

Let me start by showing the problem as it appeared in the text.

Attempt :

##{\color{red}{\large *}}## The key thing is to do the problem in a "good way", whereby I can express ##x^3+y^3=f\{(x+y), (x-y)\}## purely, with no cross terms - and that is not what I am able to do.

Below I reproduce what I could do, both methods less than satisfactory.

Method 1 (bad - brute force) : Given ##x+y=m## and ##x-y=n##, we solve them to yield ##x=\dfrac{m+n}{2}, y = \dfrac{m-n}{2}##.

The required expression ##x^3+y^3=\dfrac{1}{8}\{(m+n)^3+(m-n)^3\}=\dfrac{1}{8}(2m^3+6mn^2)=\boxed{
\dfrac{m}{4}(m^2+3n^2)}\quad{\color{green}{\Large{\checkmark}}}##, agreeing with the answer in the text shown alongside.

Method 2 (partly satisfactory) : Let's write ##\small{x^3+y^3 = (x+y)(x^2-xy+y^2)=(x+y)\{(x-y)^2+xy\}=m\left\{n^2+\dfrac{1}{4}(m^2-n^2)\right\}=\dfrac{m}{4}(4n^2+m^2-n^2)=\boxed{\dfrac{m}{4}(m^2+3n^2}}\quad{\color{green}{\Large{\checkmark}}}##

Request : Is there a neater way to solve it, as I mentioned above alongside the ##{\color{red}{\large *}}## ?

 

[Ibix]

Method 2 (partly satisfactory) : Let's write ##x^3+y^3 = (x+y)(x^2-xy+y^2)##

Sure would be nice if that second term in brackets were a perfect square. Shame about the ##xy## term - can you think of a way to express it in terms of ##x+y## and ##x-y## so you can make it into a ##\pm 2xy## in a convenient form?

 

[Ibix]

Incidentally, your solution for ##x## is upside down in method 1.

 

[brotherbobby]

Sure would be nice if that second term in brackets were a perfect square or a difference of two squares. Shame about the ##xy## term - can you think of a way to express it in terms of ##x+y## and ##x-y## so you can make it go away?

Aha I remember ##(x+y)^2-(x-y)^2=4xy\Rightarrow xy = \dfrac{1}{4}\{(x+y)^2-(x-y)^2\}##.

Worth trying (2nd method above):

##\small{x^3+y^3 = (x+y)(x^2-xy+y^2)=(x+y)\{(x-y)^2+\underline{xy}\}}##.

Replacing the underlined term :

##\small{x^3+y^3=(x+y)[(x-y)^2+\dfrac{1}{4}\{(x+y)^2-(x-y)^2\}]=m\left\{n^2+\dfrac{1}{4}(m^2-n^2)\right\}=\boxed{\dfrac{m}{4}\left(3n^2+m^2\right)}}\quad\checkmark##

 

[brotherbobby]

Incidentally, your solution for ##x## is upside down in method 1.

Thanks. Corrected.

 

[Ibix]

Yep. So if something looks nearly like something you know how to solve, hammer it into shape and see if you can deal with the leftovers. Sometimes that doesn't help, but sometimes it does.

 

[Hill]

Here is a different approach.

It is clear that the answer is a polynomial ##am^3+bm^2n+cmn^2+dn^3##. By assigning 4 convenient combinations of ##x,y## it is easy to determine ##a,b,c,d##.

For example, ##x=1,y=1## immediately gives you ##a=\frac 1 4##, and ##x=1, y=-1## gives ##d=0##.

##x=1, y=0## gives ##b+c=\frac 3 4##, and ##x=0, y=1## gives ##-b+c=\frac 3 4##, which lead to ##b=0, c=\frac 3 4##.

So, ##x^3+y^3=\frac 1 4 m^3+\frac 3 4 mn^2##.

 

[brotherbobby]

Yep. So if something looks nearly like something you know how to solve, hammer it into shape and see if you can deal with the leftovers. Sometimes that doesn't help, but sometimes it does.

Yes, I suppose it's a mistake to come on here too soon.

 

[Ibix]

Yes, I suppose it's a mistake to come on here too soon.

Possibly, although it's difficult to know when it stops being "too soon". I wasn't criticising in the post you quoted - just trying to be explicit about the general approach I used to solve the problem.

 

[brotherbobby]

Possibly, although it's difficult to know when it stops being "too soon". I wasn't criticising in the post you quoted - just trying to be explicit about the general approach I used to solve the problem.

Yes I understand. I must admit that I have often had the urge to come on here for problems which proved difficult after a few attempts. My rule of thumb is one day - which I realise is a little too soon. A better rule of thumb would be to wait a few more days and give the problem different lines of attack.

 

[Mark44]

Method 1 (bad - brute force)
Given ##x+y=m## and ##x-y=n##, we solve them to yield ##x=\dfrac{m+n}{2}, y = \dfrac{m-n}{2}##.

I don't see anything wrong with this approach, so I wouldn't describe it as bad. I think it's the most straightforward way to go about things, so I wouldn't call it brute force, either.
Once you have solved for x and y in terms of m and n, just substitute them into the ##x^3 + y^3##. expression.

 

[pbuk]

I don't see anything wrong with this approach, so I wouldn't describe it as bad. I think it's the most straightforward way to go about things, so I wouldn't call it brute force, either.

I agree, this is the methodical approach that will inevitably lead to the correct solution, you just have to be careful with your expansions and gathering terms.

The alternative of trying to find 'tricks' that will lead to a short-cut only works if you think of the right trick, and although @Hill's method works well in this case, trial and error may not always lead quickly to the right solution.

KISS.

 

[PeroK]

I don't see anything wrong with this approach, so I wouldn't describe it as bad. I think it's the most straightforward way to go about things, so I wouldn't call it brute force, either.
Once you have solved for x and y in terms of m and n, just substitute them into the ##x^3 + y^3##. expression.

Note that:
$$(m+n)^3 + (m - n)^3 = m^3 + 3m^2n + 3mn^2 + n^3 + m^3 - 3m^2n + 3mn^2 - n^3 = 2(m^3 + 3mn^2)$$Where the RHS is twice the sum of the binomial terms with even powers of ##n##.

In fact, using this method, we can write down the solution for a higher power with minimal effort:
$$x^5 + y^5 = \frac 1 {16}\big[m^5 + 10m^3n^2 + 5mn^4]$$

 

[pasmith]

Here is a different approach.

It is clear that the answer is a polynomial ##am^3+bm^2n+cmn^2+dn^3##. By assigning 4 convenient combinations of ##x,y## it is easy to determine ##a,b,c,d##.

[itex]x^3 + y^3[/itex] is symmetric, so [itex]n = x - y[/itex] can only appear raised to an even power. That immediately gives [itex]b = d = 0[/itex]. And then it follows from the coefficient of [itex]x^3[/itex] that [itex]a + c = 1[/itex] and from the coefficient of [itex]x^2y[/itex] that [itex]0 = 3a - c[/itex].

It's easier to expand [itex]mn^2[/itex] if you write it as [itex](x - y)(x^2 - y^2)[/itex].

 

[brotherbobby]

@PeroK's observations (post #13) can be extended to even powers also - how would, for example, ##x^4+y^4## look? [Given ##x+y=m,\, x-y=n##]

The R.H.S. of the expression would be twice of the sum of the binomial coefficients with even powers of ##m##.

##x^4+y^4 = \dfrac{1}{2^3}\left( m^4+6m^2n^2+n^4 \right)##.

Turns out, the brute force method of doing this problem, wherein we rewrite ##x,y## in terms of ##m,n## is not a bad one after all.

 

[PeroK]

Turns out, the brute force method of doing this problem, wherein we rewrite ##x,y## in terms of ##m,n## is not a bad one after all.

I saw my method more as jujitsi than brute force!

 

[brotherbobby]

jujutsi

Jujitsu you mean.

Yes there's an element of intelligence about the method. Nonetheless, asked some different and more complicated expression of the form of ##f(x,y)=?,## we can proceed with ##\small{x+y=m, x-y=n\Rightarrow x=\dfrac{m+n}{2}, y=\dfrac{m-n}{2}}## and plug in the variables ##m,n## to replace ##x,y##. If you tried it that way, the "brute force" character of the situation would become apparent to you.

On the other hand, the book I am following [Hall and Knights Elementary Algebra (1888)], the authors suggest a nicer way - express the function ##f(x,y)=g\{ (x+y), (x-y)\}##, via algebra, where ##g## will be (obviously) a different function. If you do that, the result is straightforward ##-## just replace ##x+y=m, x-y=n## wherever you find them in ##g##.

I copy and paste an example from the text. [By the way, the word should be show in the text below and not "shew". "Shew" being repeated throughout the text, I think this is how they spoke and wrote English in old England. Also note the error in the 2nd line of the solution ##-## it should read ##x^4## and not ##x^2##.]

 

[PeroK]

I think the binomial method is quite general and easy to remember:
$$x^2 + y^2 = \frac 1 2(a^2 + b^2)$$$$x^4 + y^4 = \frac 1 8(a^4 + 6a^2b^2 + b^4)$$$$xy = \frac 1 4(a^2 - b^2)$$$$x^2y^2 = \frac 1 {16}(a^4 -2a^2b^2 + b^4)$$Then, we have:
$$4(x^4 - 6x^2y^2 +y^4) = 4(x^4 +y^4) -24x^2y^2$$$$ = \frac 1 2(a^4 + 6a^2b^2 + b^4) - \frac 3 2(a^4 -2a^2b^2 + b^4)$$$$=6a^2b^2 - a^4 - b^4$$