1 problem found
Show that \(x^3-3xbc + b^3 + c^3\) can be written in the form \(\left( x+ b+ c \right) {\rm Q}( x)\), where \({\rm Q}( x )\) is a quadratic expression. Show that \(2{\rm Q }( x )\) can be written as the sum of three expressions, each of which is a perfect square. It is given that the equations \(ay^2 + by + c =0\) and \(by^2 + cy + a = 0\) have a common root \(k\). The coefficients \(a\), \(b\) and \(c\) are real, \(a\) and \(b\) are both non-zero, and \(ac \neq b^2\). Show that \[ \left( ac - b^2 \right) k = bc - a^2 \] and determine a similar expression involving \(k^2\). Hence show that \[ \left( ac - b^2 \right) \left(ab-c^2 \right) = \left( bc - a^2 \right)^2 \] and that \( a^3 -3abc + b^3 +c^3 = 0\,\). Deduce that either \(k=1\) or the two equations are identical.
Solution: \begin{align*} && x^3 - 3xbc+b^3 + c^3 &= (x+b+c)(x^2-x(b+c)+b^2+c^2-bc) \\ &&&= \tfrac12(x+b+c)((x-b)^2+(x-c)^2+(b-c)^2) \\ \end{align*} We must have: \begin{align*} && 0 &= ak^2 + bk+c \tag{1}\\ &&0 &= bk^2+ck+a \tag{2}\\ b*(1)&& 0 &= abk^2 + b^2k+cb \\ a*(2)&& 0 &= abk^2 + ack + a^2 \\ \Rightarrow && 0 &= k(ac-b^2)+a^2-bc \\ \Rightarrow && (ac-b^2)k &= bc-a^2 \\ \\ c*(1) && 0 &= ack^2+bck+c^2 \\ b*(2) && 0 &= b^2k^2+bck+ab \\ \Rightarrow && 0 &= (ac-b^2)k^2 +c^2-ab \\ \Rightarrow && (ac-b^2)k^2 &= ab-c^2 \\ \\ \Rightarrow && \frac{ab-c^2}{ac-b^2} &= k^2 = \left (\frac{bc-a^2}{ac-b^2} \right)^2 \\ \Rightarrow && (ab-c^2)(ac-b^2) &= (bc-a^2)^2 \\ \Rightarrow && a^2bc - ab^3-ac^3+b^2c^2 &= b^2c^2-2a^2bc+a^4 \\ \Rightarrow && 0 &= a^4+ab^3+ac^3-3a^2bc \\ &&&= a(a^3+b^3+c^3-3abc) \\ \underbrace{\Rightarrow}_{a \neq 0} && 0 &= a^3+b^3+c^3-3abc \\ &&&= (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \end{align*} Therefore \(a+b+c = 0\). (Since otherwise \(a=b=c\) but \(ac \neq b^2\)). This means \(1\) is a root of our equations. Therefore, either \(k = 1\) or they have both roots in common, ie they are the same equation up to a scalar factor. ie \(b = la, c = lb, a= lc \Rightarrow l^3 = 1 \Rightarrow l = 1\). Therefore, they are the same equation.