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.