if alpha, beta and gamma are zeroes of 4x^3-32x^2+25x+21=0, then value of
[alpha^3/{(alpha-beta)(alpha-gamma)}]
+
[beta^3/{(beta-alpha)(beta-gamma)}]
+
[gamma^3/{(gamma-alpha)(gamma-beta)}]
is?
Looking at the 21, I suspect that roots could be ±1, ±3, ±7
after a few tries, when x = 7
4(7)^3-32(7)x^2+25(7)x+21=0
So I did synthetic division and
4x^3-32x^2+25x+21=0
(x-7)(4x^2 - 4x - 3) = 0
(x-7)(2x + 1)(2x - 3) = 0
so your zeros are 7, -1/2 and 3/2
Now perform that gibberish on them that you stated as:
[alpha^3/{(alpha-beta)(alpha-gamma)}]
+
[beta^3/{(beta-alpha)(beta-gamma)}]
+
[gamma^3/{(gamma-alpha)(gamma-beta)}]
Well, buckle up because we're about to take a trip down math-funny lane, or as I like to call it, "Algebra Circus." So, let's get started.
First, let's talk about the denominator. We have terms like (alpha-beta) and (alpha-gamma) in the expression. Now, if you ask me, it's quite amusing how these terms seem to be feuding with each other. It's like they have some sort of "mathematical rivalry" going on. Perhaps there's a "math gang war" happening between these terms. Who knows?
Now, let's think about the numerator. We have terms like alpha^3, beta^3, and gamma^3. These terms are trying to show off their "cubic power." It's like they're flexing their mathematical muscles, saying, "Hey, look at us! We can raise ourselves to the power of three!" They must be feeling pretty confident, huh?
So, what do we get when we combine these numerator and denominator expressions? Well, let's take a look:
[alpha^3/{(alpha-beta)(alpha-gamma)}]
+ [beta^3/{(beta-alpha)(beta-gamma)}]
+ [gamma^3/{(gamma-alpha)(gamma-beta)}]
It appears that these expressions are having a "mathematical potluck party." Each term brings its own cubic power to the table, hoping to make the sum as entertaining as possible. It's like a "mathematical talent show" happening right before our eyes.
But, my dear friend, the value of this expression is not just all fun and games. To actually determine the value, we need to use Vieta's formulas. However, I hope my little excursion into the world of "math humor" made you smile. After all, laughter is always the best solution, even when it comes to math problems.
To find the value of the expression given, we can first calculate the individual terms separately and then add them together.
Given the equation: 4x^3 - 32x^2 + 25x + 21 = 0
Let's find the sum of cubes for each term:
For alpha:
Alpha^3 = alpha * alpha * alpha
For the denominator:
(alpha - beta) * (alpha - gamma) = alpha^2 - alpha*beta - alpha*gamma + beta*gamma
Therefore, the first term becomes alpha^3 / (alpha^2 - alpha*beta - alpha*gamma + beta*gamma)
Similarly for beta and gamma:
For beta:
Beta^3 = beta * beta * beta
For the denominator:
(beta - alpha) * (beta - gamma) = beta^2 - beta*alpha - beta*gamma + alpha*gamma
Therefore, the second term becomes beta^3 / (beta^2 - beta*alpha - beta*gamma + alpha*gamma)
For gamma:
Gamma^3 = gamma * gamma * gamma
For the denominator:
(gamma - alpha) * (gamma - beta) = gamma^2 - gamma*alpha - gamma*beta + alpha*beta
Therefore, the third term becomes gamma^3 / (gamma^2 - gamma*alpha - gamma*beta + alpha*beta)
Now, we can add these terms together:
(alpha^3 / (alpha^2 - alpha*beta - alpha*gamma + beta*gamma)) +
(beta^3 / (beta^2 - beta*alpha - beta*gamma + alpha*gamma)) +
(gamma^3 / (gamma^2 - gamma*alpha - gamma*beta + alpha*beta))
Simplifying and factoring out common factors:
[alpha^3 + beta^3 + gamma^3] / [(alpha^2 - alpha*beta - alpha*gamma + beta*gamma) +
(beta^2 - beta*alpha - beta*gamma + alpha*gamma) +
(gamma^2 - gamma*alpha - gamma*beta + alpha*beta)]
Expanding and simplifying further:
[(alpha + beta + gamma) * (alpha^2 - alpha*beta - alpha*gamma + beta*gamma)] /
[(alpha^2 - alpha*beta - alpha*gamma + beta*gamma) +
(beta^2 - beta*alpha - beta*gamma + alpha*gamma) +
(gamma^2 - gamma*alpha - gamma*beta + alpha*beta)]
Notice that the numerator and denominator have a common factor: (alpha^2 - alpha*beta - alpha*gamma + beta*gamma)
Cancelling out the common factor, we are left with:
(alpha + beta + gamma) / (alpha + beta + gamma)
Simplifying further:
1
Therefore, the value of the given expression is 1.
To find the value of the given expression, we can use Vieta's formulas, which relate the coefficients of a polynomial to its roots.
In this case, the given equation is 4x^3 - 32x^2 + 25x + 21 = 0. Vieta's formulas state that for a monic polynomial of the form ax^3 + bx^2 + cx + d = 0, the sum of the roots is -b/a, the sum of the product of the roots taken two at a time is c/a, and the product of the roots is -d/a.
So, let's calculate the sum of the roots (alpha, beta, and gamma):
Sum of the roots = (-coeffiecient of x^2) / (coefficient of x^3)
= -(-32) / 4
= 8
Now, let's calculate the product of the roots:
Product of the roots = (-constant term) / (coefficient of x^3)
= -21 / 4
= -5.25
Next, let's calculate the expression alpha^3 / ((alpha-beta)(alpha-gamma)).
Using Vieta's formulas, we have (alpha-beta)(alpha-gamma) = alpha^2 - (alpha)(beta+gamma) + beta*gamma.
Plugging in the values we found earlier:
(alpha-beta)(alpha-gamma) = alpha^2 - (alpha)(8-alpha) + (beta*gamma)
= alpha^2 - 8alpha + alpha^2 + 5.25 (since beta+gamma = 8 and beta*gamma = -5.25)
= 2alpha^2 - 8alpha + 5.25
Now, let's simplify the expression alpha^3 / ((alpha-beta)(alpha-gamma)):
alpha^3 / ((alpha-beta)(alpha-gamma))
= alpha^3 / (2alpha^2 - 8alpha + 5.25)
Simplifying further:
= alpha / (2alpha - 8 + 5.25/alpha)
= (1 / 2 - 8/alpha + 5.25/alpha^2)
Using the same process, we can calculate the expressions for beta^3 / ((beta-alpha)(beta-gamma)) and gamma^3 / ((gamma-alpha)(gamma-beta)).
Finally, we will substitute the values of alpha, beta, and gamma into the expressions we found, and then add them together to get the final result.