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If $ab = 2$ and $a+b = 5$ then calculate the value of $a^4+b^4$

My approach: $$a^4+b^4 = (a+b)^4-4a^3b-6a^2b^2-4ab^3$$ $$=(5)^4-6(ab)^2-4ab.a^2-4ab.b^2$$ $$=(5)^4-6(24)-4ab(a^2-b^2)$$ $$=(5)^4-6(24)-8(a+b)(a-b)$$ $$=(5)^4-6(24)-8(5)(a-b)$$ I am a little stuck now and any help will be appreciated.

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2  
Check your factorisation $-4ab\cdot a^2-4ab \cdot b^2 = -4ab(a^2+b^2)$. –  Macavity May 2 at 10:59
    
Another possible way of solving this could be using the following factorization: $$a^4+b^4=(a+b)(a^3+b^3)-ab(a^2+b^2)$$ –  mathh May 2 at 11:35

6 Answers 6

up vote 3 down vote accepted

\begin{align} a^4+b^4 &= (a+b)^4-4a^3b-6a^2b^2-4ab^3 \\&= (a+b)^{4} - 4 ab (a^{2} + b^{2}) - 6 (ab)^{2} \\&= (a+b)^{4} - 4 ab ((a+b)^{2} - 2 ab) - 6 (ab)^{2} \\&= (a+b)^{4} - 4 ab (a+b)^{2} + 2 (ab)^ {2}. \end{align}

As noted in a comment, you made a sign error (it happens) in your calculations, otherwise you would have got here yourself.

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Thanks! Noticed my mistake. –  Aspiring Mathlete May 2 at 12:10

You could start with $a+b$ and $ab$,

  1. Note first $(a+b)^2 - 2ab = a^2+ b^2 $

  2. Note that $(a^2+b^2)^2 - 2(ab)^2 = a^4 + b^4$

As an asside, i use this set of numbers to factorise $b^n-a^n$. Part of the process involves creating a sequence $T_n = a^n+b^n$. You find that $T_{n+1}=(a+b)T_n - ab T_{n-1}$

Applying $T_0=2$ and $T_1$=5 we get this series for the sum of $a^n+b^n$ for n=0 t0 6. The iteration is t(n+1)=5t(n)-2t(n-1).

      0    1    2    3    4     5      6
      2    5   21   95  433  2070   9484

Doing it this way, allows one to evaluate symmetric equations (ie $f(a,b)=f(b,a)), very quickly. You start at the middle, and work outwards, multiplying the result by ab, before adding the next term.

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There is a general recursive identity for such questions: let $f(n) = a^n + b^n$. Then $$f(n+1) = f(1)f(n) - ab f(n-1).$$ Note that $f(0) = 2$ provided $ab \ne 0$. Then in your particular case, we wish to find $f(4)$, where $$f(n+1) = 5f(n) - 2f(n-1).$$ With starting values $f(0) = 2$ and $f(1) = 5$, we easily compute $f(2) = 5(5)-2(2) = 21$, $f(3) = 5(21) - 2(5) = 95$, $f(4) = 5(95) - 2(21) = 433$. The advantage of this approach is that it can be used to compute sums of higher powers quite easily. It also leads to a general solution via the solution of the associated linear recurrence; e.g., with generating functions.

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An easier approach: Since $ab = 2$, then

$$ (a+b)^2 = 25 $$. Hence

$$ a^2 + b^2 + 2ab = 25 \iff a^2 + b^2 = 25 - 2 ab = 21 $$

Hence,

$$ (21)^2 = (a^2 + b^2)^2 = a^4 + b^4 + 2 a^2 b^2 = a^4 + b^4 + 2(2)^2 $$

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Another totally different approach is tho solve for $a$ and $b$ using : $$ab = 2, \qquad a+b = 5$$ This is like solving a second order polynomial $(x+a)(x+b) = x^2 + (a+b)x+ab$ where the roots are $-a$ and $-b$.

So you could solve the polynomial $x^2 + 5x + 2 = 0$, deduce $a$ and $b$ and calculate $a^4+b^4$.

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I might be totally wrong, but it seems to me that the whole point of the exercise is precisely to avoid computing the exact values of $a, b$, which involve a square root, and use only their sum and product, which are integers. –  Andreas Caranti May 2 at 11:04
    
@AndreasCaranti: Possibly indeed but the OP didn't specify so I thought I'll give another option. –  user88595 May 2 at 11:06

Use $a+b=5$ to write $b$ in terms of $a$: $$b=5-a$$ Then substitute this for b in the other equation and solve for $a$: $$a(\overbrace{5-a}^{b}) =2\iff 5a-a^2=2\iff a^2-5a+2=0\iff a=\frac52\pm\frac12\sqrt{17}$$ (using the quadratic formula). Since the original equations were symmetric in $a$ and $b$ (you could have solved for $b$ instead and arrived at exactly the same values), these two numbers (the $\pm$) are the values of $a$ and $b$: $$a=\frac52\pm\frac12\sqrt{17} \textrm{ and } b=\frac52\mp\frac12\sqrt{17}$$

Since this is homework, I will leave it up to you to compute that $$a^4+b^4 = \boxed{433}$$ given these values of $ a$ and $b$.

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