Question 1 Two numbers are such that the sum of their cubes is 14 and the sum of their squares is 6. Find the sum of the two numbers.
I did $a^2+b^2=6$ and $a^3+b^3=14$ Find $a$ and $b$, two numbers. but got lost when trying to algebraicly solve it.
Thank you, Any help is appreciated
There are lots of ways of solving this. One is to note that $a$ and $b$ are the roots of the quadratic equation $$0=(x-a)(x-b)=x^2-(a+b)x+ab=x^2-px+q$$ Where we use $p=a+b$ and $q=ab$ for the unknown coefficients.
Then we have $a^2-pa+q=b^2-pb+q=0$ and adding the equations we obtain $$(a^2+b^2)-p(a+b)+2q=0$$ which becomes$$6-p^2+2q=0$$
We also have (after multiplying the quadratic equation by $x$ and substituting $a,b$ and adding):
$$(a^3+b^3)-p(a^2+b^2)+q(a+b)=0$$which becomes $$14-6p+pq=0=28-12p+2pq=28-12p+p(p^2-6)$$ (substituting for $2q$ from the first equation)
Which gives the cubic for $p$ $$p^3-18p+28=0$$ This can be solved using standard methods. Pretty much any method will come down to solving an equivalent cubic.
Note that $p=a+b$ is what you are asked to find.
The problem doesn't specify any restriction on the numbers $a$ and $b$, so we're going to assume they are complex. Let $S=a+b$ denote the sum. Then
$$S^2=(a+b)^2=a^2+2ab+b^2=6+2ab$$
and
$$S^3=(a+b)^3=a^3+3a^2b+3ab^2+b^3=14+3abS$$
Rewriting the first of these as $2ab=S^2-6$ and multiplying both sides of the second by $2$ gives
$$2S^3=28+3(2ab)S=28+3(S^2-6)S=28+3S^3-18S$$ or
$$S^3-18S+28=(S-2)(S^2+2S-14)=0$$
The possible values of $S$ are thus $2$, $-1+\sqrt{15}$, and $-1-\sqrt{15}$.
If you care to chase down the actual values of $a$ and $b$, the first two of these give real values while the third gives a pair of complex conjugates. In particular, $S=2$ leads to $1\pm\sqrt2$ for $a$ and $b$.
When numbers are so small you can proceed by attempt. However $a^2+b^2=6$ doesn't have no integer solution: wlog you can take $a,b\geq0$.
So try: $a=0\Rightarrow b=\sqrt6$ (not valid in $\mathbb Z$).
$a=1\Rightarrow b=\sqrt5$ (not valid as above)
$a=2\Rightarrow b=\sqrt2$ (idem).
$a\geq3\Rightarrow b^2<0$.
But maybe you mean to find $a,b\in\mathbb R$.
In this case just notice that $14=a^3+b^3=x(6-y)$ and $6=a^2+b^2=x^2-2y$, where $x=a+b$ and $y=ab$. Solving the system you'll find $x=\frac{14}{6-y}$ and $y^3-9y^2+10=0$, which is solvable using the well known formulas for the 3rd degree polynomials.
There's another way to get to answer if we take $a=c+d$ and $b=c-d$, then using binomial expansions we have $$ a^2+b^2=2(c^2+d^2)=6\\ a^3+b^3=2c^3+6cd^2=14 $$
We have $d^2=3-c^2$ from first equation and using this in second equation $$ c(9-2c^2)=7 $$
We can easily see that one of the solutions is c=1.
