Question

If $ax^2+bx+c=0$ and $bx^2+cx+a=0$ have a common root, and a, b, c are non-zero, then $(\frac{a^3+b^3+c^3}{abc})$

Answer 1

3

Answer 2

The condition for two quadratic equations $A_1x^2+B_1x+C_1=0$ and $A_2x^2+B_2x+C_2=0$ to have a common root is $(A_1C_2 - A_2C_1)^2 = (A_1B_2 - A_2B_1)(B_1C_2 - B_2C_1)$. For the given equations $ax^2+bx+c=0$ and $bx^2+cx+a=0$, we have $A_1=a, B_1=b, C_1=c$ and $A_2=b, B_2=c, C_2=a$. Substituting these values into the condition: $(a \cdot a - b \cdot c)^2 = (a \cdot c - b \cdot b)(b \cdot a - c \cdot c)$ $(a^2 - bc)^2 = (ac - b^2)(ab - c^2)$ Expanding both sides: $a^4 - 2a^2bc + b^2c^2 = a^2bc - ac^3 - ab^3 + b^2c^2$ $a^4 - 2a^2bc = a^2bc - ac^3 - ab^3$ $a^4 + ab^3 + ac^3 = 3a^2bc$ Since $a \neq 0$, divide by $a$: $a^3 + b^3 + c^3 = 3abc$ Therefore, $\frac{a^3+b^3+c^3}{abc} = \frac{3abc}{abc} = 3$.