Question

the sum of the roots of the equation 5x²+(p+q+r)x+pqr=0.what is the value of (p³+q³+ŕ³)​

Answer 1

Answer:

Step-by-step explanation:

It is given that

P+Q+R=0

We have to prove that :  

We know that

cubing both sides

=> (P+Q)³ = (-R)³

=> P³+Q³+ 3PQ(P+Q) = -R³

=> P³+Q³+3PQ (-R) = -R³. (from equation 1)

=> P³+Q³-3PQR = -R³

=> P³+Q³+R³ = 3PQR

Hence proved.

Answer 2

Answer: 125p³ + 125q³ + 125r³

Step-by-step explanation:

We can use Vieta's formulas to solve this problem. Vieta's formulas relate the coefficients of a polynomial to its roots. Specifically, for a quadratic equation of the form ax²+bx+c=0 with roots r₁ and r₂, the sum of the roots is -b/a, and the product of the roots is c/a.

In this case, we have the quadratic equation 5x² + (p+q+r)x + pqr = 0. Let the roots of this equation be r₁ and r₂. Then, by Vieta's formulas, we have:

r₁ + r₂ = -(p+q+r)/5

and

r₁r₂ = pqr/5

We want to find the value of p³+q³+r³. To do this, we can use the identity:

p³ + q³ + r³ = (p+q+r)³ - 3(p+q+r)(pq+qr+rp) + 3pqr

We already know the value of p+q+r from the first Vieta formula. To find the value of pq+qr+rp, we can square the first Vieta formula and simplify:

(r₁ + r₂)² = (r₁)² + 2r₁r₂ + (r₂)²

(-1)(p+q+r)²/25 = r₁² + 2r₁r₂ + r₂²

-5(p+q+r)² = 25r₁² + 50r₁r₂ + 25r₂²

-(p+q+r)² = 5r₁² + 10r₁r₂ + 5r₂²

-(p+q+r)² = 5(r₁ + r₂)² - 10r₁r₂

-(p+q+r)² = 5(-b/a)² - 10(c/a)

-(p+q+r)² = 5((p+q+r)/5)² - 10(pqr/5)

-(p+q+r)² = (p+q+r)²/5 - 2pqr

Therefore,

pq+qr+rp = -(p+q+r)²/5 + pqr

Substituting into the identity for p³+q³+r³, we get:

p³ + q³ + r³ = (p+q+r)³ - 3(p+q+r)(-(p+q+r)²/5 + pqr) + 3pqr

Simplifying, we get:

p³ + q³ + r³ = (p+q+r)³ + 3(p+q+r)²/5 - 6pqr

Substituting in the values we found earlier, we get:

p³ + q³ + r³ = (-(p+q+r)/5)³ + 3(-(p+q+r)/5)²/5 - 6(pqr/5)

Simplifying further, we get:

p³ + q³ + r³ = -(p+q+r)³/125 + 3(p+q+r)²/25 - 6pqr/5

Substituting in the values of r₁+r₂ and r₁r₂ that we found earlier, we get:

p³ + q³ + r³ = -(-(p+q+r)/5)³/125 + 3(-(p+q+r)/5)²/25 - 6(pqr/5)

Simplifying further and multiplying through by 125, we get:

125p³ + 125q³ + 125r³

Learn more about Vieta's formulas :

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