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Q-p-q-r-are-roots-of-x-3-7x-2-4-x-x-2-find-the-value-of-K-1-qr-1-3-1-pr-1-3-1-pq-1-3-

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Question Number 174127 by mnjuly1970 last updated on 26/Jul/22
Q: p , q , r are roots of x^( 3) −7x^( 2) =(4−x)(x+2) find the value of : K = (1/( ((qr))^(1/3) )) + (1/( ((pr))^(1/3) )) + (1/( ((pq))^(1/3) )) = ?
Q:p,q,rarerootsofx3−7x2=(4−x)(x+2)findthevalueof:K=1qr3+1pr3+1pq3=?
Answered by Rasheed.Sindhi last updated on 26/Jul/22
x^3 +5x^2 =8x−15−x^2 x^3 +6x^2 −8x+15=0 p+q+r=−6 , pq+qr+rp=−8 , pqr=−15 K = (((p)^(1/3) +(q)^(1/3) +(r)^(1/3) )/( ((pqr))^(1/3) )) ⇒(p)^(1/3) +(q)^(1/3) +(r)^(1/3) = ((pqr))^(1/3) K K^3 = ((((p)^(1/3) +(q)^(1/3) +(r)^(1/3) )/( ((pqr))^(1/3) )))^3 (a+b+c)^3 =a^3 +b^3 +c^3 +3(ab+bc+ca)(a+b+c)−3abc =((p+q+r+3(((pq))^(1/3) +((qr))^(1/3) +((rp))^(1/3) )((p)^(1/3) +(q)^(1/3) +(r)^(1/3) )−3((pqr))^(1/3) )/(pqr)) =((p+q+r+3(((pq))^(1/3) +((qr))^(1/3) +((rp))^(1/3) )(((pqr))^(1/3) K )−3((pqr))^(1/3) )/(pqr)) =((p+q+r+3(((((pqr))^(1/3) )/( (r)^(1/3) ))+((((pqr))^(1/3) )/( (p)^(1/3) )) +((((pqr))^(1/3) )/( (q)^(1/3) )) )(((pqr))^(1/3) K )−3((pqr))^(1/3) )/(pqr)) =((p+q+r+3(((pqr))^(1/3) )^2 (((1 )/( (r)^(1/3) ))+((1 )/( (p)^(1/3) )) +((1 )/( (q)^(1/3) )) )^★ (K )−3((pqr))^(1/3) )/(pqr)) K^3 (pqr)=p+q+r+3(((pqr))^(1/3) )^2 (A )(K )−3((pqr))^(1/3) determinant (((K^3 (pqr)=p+q+r+3(((pqr))^(1/3) )^2 (A )(K )−3((pqr))^(1/3) )))...(I) ^★ A=((1 )/( (r)^(1/3) ))+((1 )/( (p)^(1/3) )) +((1 )/( (q)^(1/3) )) A^3 =(1/r)+(1/p)+(1/q)+3((1/( ((pq))^(1/3) ))+(1/( ((qr))^(1/3) ))+(1/( ((rp))^(1/3) )))(a)−(3/( ((pqr))^(1/3) )) A^3 =(1/r)+(1/p)+(1/q)+3(K)(A)−(3/( ((pqr))^(1/3) )) A^3 =((pq+qr+rp)/(pqr))+3(K)(A)−(3/( ((pqr))^(1/3) )) =((−8)/(−15))+3KA−(3/( ((−15))^(1/3) )) K=((A^3 −(8/(15))+(3/( ((−15))^(1/3) )))/(3A)) determinant (((K=((A^3 −(8/(15))+(3/( ((−15))^(1/3) )))/(3A)))))....(II) Solving (I) & (II) (eqns in boxes) simultaneously we can determine K Too complicated....
x3+5x2=8x−15−x2x3+6x2−8x+15=0p+q+r=−6,pq+qr+rp=−8,pqr=−15K=p3+q3+r3pqr3⇒p3+q3+r3=pqr3KK3=(p3+q3+r3pqr3)3(a+b+c)3=a3+b3+c3+3(ab+bc+ca)(a+b+c)−3abc=p+q+r+3(pq3+qr3+rp3)(p3+q3+r3)−3pqr3pqr=p+q+r+3(pq3+qr3+rp3)(pqr3K)−3pqr3pqr=p+q+r+3(pqr3r3+pqr3p3+pqr3q3)(pqr3K)−3pqr3pqr=p+q+r+3(pqr3)2(1r3+1p3+1q3)★(K)−3pqr3pqrK3(pqr)=p+q+r+3(pqr3)2(A)(K)−3pqr3K3(pqr)=p+q+r+3(pqr3)2(A)(K)−3pqr3…(I)★A=1r3+1p3+1q3A3=1r+1p+1q+3(1pq3+1qr3+1rp3)(a)−3pqr3A3=1r+1p+1q+3(K)(A)−3pqr3A3=pq+qr+rppqr+3(K)(A)−3pqr3=−8−15+3KA−3−153K=A3−815+3−1533AK=A3−815+3−1533A….(II)Solving(I)&(II)(eqnsinboxes)simultaneouslywecandetermineKToocomplicated….

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