From Quadratic To Cubic

Let $g(x)=f(x)-x$ . We see that $g(a)=0$ , $g(b)=0$ , and $g(a+b)=0$ . Therefore $g(x)=A(x-a)(x-b)(x-(a+b))$ for some constant $A$ . This means $f(x)=A(x-a)(x-b)(x-(a+b))+x$ .

First, note that $(x-a)(x-b)=x^2-x-4$ . This means $f(x)=A(x^2-x-4)(x-(a+b))+x$ .

Also, note that by Vietas, $a+b=1$ and $ab=-4$ . This means that we can further simplify $f(x)=A(x^2-x-4)(x-1)+x$ , and that $f(-4)=-84$ .

We can now solve for $A$ : $f(-4)=A((-4)^2-(-4)-4)((-4)-1)+(-4)=-84$ , and simplifying gives $A=1$ .

Now we have the function $f(x)=(x^2-x-4)(x-1)+x$ . Plugging in $6$ for $x$ gives $f(6)=\boxed{136}$ , and we are done.

My solution was too long. By the way, here is it:

$a$ and $b$ are the roots of ${ x }^{ 2 }-x-4=0$ . Then, by Vieta´s , we have:

$a+b=1$ and $ab = -4$

Let $f(x)=m{ x }^{ 3 }+n{ x }^{ 2 }+px+q$

We see that:

$f(a)=m{ a }^{ 3 }+n{ a }^{ 2 }+pa+q = a$

$f(b)=m{ b }^{ 3 }+n{ b }^{ 2 }+pb+q = b$

$f(a+b)= f(1) = m + n + p + q = 1$

$f(ab)= f(-4) = -64m + 16n - 4p + q = -84$

We know that:

${ a }^{ 2 }+{ b }^{ 2 }={ (a+b) }^{ 2 }-2ab={ 1 }^{ 2 }-2\cdot (-4)=9$

and

${ a }^{ 3 }+{ b }^{ 3 }=(a+b)({ a }^{ 2 }-ab+{ b }^{ 2 })=(1)(9-(-4))=13$

Then, we can produce:

$f(a) + f(b) = m({ a }^{ 3 }+{ b }^{ 3 })+n({ a }^{ 2 }+{ b }^{ 2 })+p(a+b)+2q = a+b$

$\Rightarrow 13m+9n+p+2q = 1$

We also know that:

${ a }^{ 2 }-{ b }^{ 2 }={ (a+b) }(a-b)={ 1 }\cdot(a-b)=a-b$

and

${ a }^{ 3 }-{ b }^{ 3 }=(a-b)({ a }^{ 2 }+ab+{ b }^{ 2 })=(a-b)(9+(-4))=5(a-b)$

Then, we can produce:

$\quad f(a) - f(b) = m({ a }^{ 3 }-{ b }^{ 3 })+n({ a }^{ 2 }-{ b }^{ 2 })+p(a-b) = a-b$

$\Rightarrow 5m(a-b) + n(a-b)+p(a-b)=a-b$

dividing all by $(a-b)$ we get: $5m + n + p = 1$

Now we have a linear system of 4 equations in the variables m, n, p and q:

$\begin{cases} m+n+p+q=1 \\ 5m+n+p=1 \\ 13m+9n+p+2q=1 \\ -64m+16n-4p+q=-84 \end{cases}$

solving it we get: $\quad m = 1, n = -2, p = -2$ and $q = 4$ .

Thus, $f(x)={ x }^{ 3 }-2{ x }^{ 2 }-2x+4$

and then,

$f(6) = 216-72-12+4 = \boxed{136}$ .

Find the value of a and b using x^2-x-4=0. We can assume the function f(x)=(x-a) (x-b) (x-(a+b)) k+x The above function is a cubic and satisfies the given conditions by putting x=a, b,(a+b). Given f(a b)=-84, Using this we can obtain the value of the constant k, which turns out to be 1. Therefore we have the function f(x). Find f(6) by substiting x=6. f(6) turns out to be 136

Since f(a) = a, f(b) = b, f(a+b) = a+b, we can say that a, b and (a+b) are roots of the polynomial function g(x) = f(x) - x = 0. So, g(x) = x^3 - (a+b+(a+b))x^2 + (ab + a(a+b) + b(a+b))x - ab(a+b). Hence, g(x) = x^3 - 2(a+b)x^2 + ((a+b)^2 + ab)x - (ab)(a+b). a and b are roots of x^2 - x -4 = 0. Hence, a+b = 1, ab = -4. Hence, g(x) = x^3 - 2x^2 - 3x +4. g(x) = f(x) - x. Hence f(x) = g(x) + x. Therefore, f(x) = x^3 - 2x^2 - 2x +4. Hence, f(6) = 136.