If ax²+bx+c=0 and bx²+cx+a=0 have common root a,b,c are non zero real numbers, then find the value of a³+b³+c³/ abc

To find the value of (a³+b³+c³) / (abc), let's consider the given equations:

Equation 1: ax² + bx + c = 0
Equation 2: bx² + cx + a = 0

Since both equations have a common root, let's consider the common root as "k".

Substituting "k" into Equation 1, we get:
ak² + bk + c = 0 ---(1)

Substituting "k" into Equation 2, we get:
bk² + ck + a = 0 ---(2)

Now, let's multiply Equation 1 by "b" and Equation 2 by "a":

b(ak² + bk + c) = 0
a(bk² + ck + a) = 0

Expanding both equations:

abk² + b²k + bc = 0 ---(3)
abk² + ack + a² = 0 ---(4)

Since Equations (3) and (4) are equal to zero and both contain the term "abk²", we can equate them to solve for k²:

abk² + b²k + bc = abk² + ack + a²

Subtracting abk² from both sides:

b²k + bc = ack + a²

Rearranging the terms:

b²k - ack = a² - bc

Factoring out "k" from the left-hand side and grouping the terms on the right-hand side:

k(b² - ac) = a² - bc

Since a, b, and c are nonzero real numbers, we can divide both sides of the equation by (b² - ac):

k = (a² - bc) / (b² - ac) ---(5)

Now, we can substitute the value of "k" from Equation (5) into the equations:

a³ = ak² = a(a² - bc) / (b² - ac)
b³ = bk² = b(a² - bc) / (b² - ac)
c³ = ck² = c(a² - bc) / (b² - ac)

To find the value of a³ + b³ + c³, we can add all three equations:

a³ + b³ + c³ = a(a² - bc) / (b² - ac) + b(a² - bc) / (b² - ac) + c(a² - bc) / (b² - ac)

Factoring out (a² - bc):

a³ + b³ + c³ = (a + b + c)(a² - bc) / (b² - ac)

Since we know that a + b + c = 0 (since they are the roots of both equations), the equation simplifies further:

a³ + b³ + c³ = 0(a² - bc) / (b² - ac)

a³ + b³ + c³ = 0

Therefore, the value of (a³ + b³ + c³) / (abc) is 0.