Let `a ,b ,c` be real numbers with `a!=0a n dl e talpha,beta` be the roots of the equation `a x^2+b x+c=0.` Express the roots of `a^3x^2+a b c x+c^3=0` in terms of `alpha,betadot`

Answer

`x=alpha^(2)beta,alphabeta^(2)`

Explanation

Since, `ax^(2)+bx+c=0` has roots `alpha and beta.`
`impliesalpha+beta=-b//a`
and ` alphabeta=c//a`
`Now, a^(3)x^(2)+abcx+c^(3)=0" "...(i)`
On dividing the equation by `c^(2),` we get
`(a^(3))/(c^(2))x^(2)(abcx)/(c^(2))+(c^(3))/(c^(2))=0`
`impliesa((ax)/(c))^(2)+b((ax)/(c))+c=0`
`impliesx=c/aalpha,c/abeta` are the roots
`implies x=alphabeta, alpha beta beta` are the roots
`impliesx=alpha^(2)beta,alphabeta^(2)` are teh roots
Divide the Eq. (i) by `a^(3),` we get
`x^(2)+b/c.c/ax+((c)/(a))^(3)=0`
`impliesx^(2)-(alpha+beta)(alphabeta)x+(alphabeta)^(3)=0`
`impliesx^(2)-alpha^(2)betax-alpha beta ^(2)x+(alphabeta)^(2)=0`
`impliesx(x-alpha^(2)beta)-alphabeta^(2)(x-alphabeta^(2))=0`
`implies (x-alpha^(2)beta)(x-alphabeta^(2))=0`
`impliesx=alpha^(2)beta,alpha beta^(2)` which is the required answer.
Alternate Solution
Since, `a^(3)x^(2)+abcx+c^(3)=0`
`impliesx=(-abcpmsqrt((abc)^(2)-4.a^(3).c^(3)))/(2a^(3))`
`impliesx=(-(b//a)(c//a)pmsqrt((b//c)^(2)(c//a)^(2)-4(c//a)^(3)))/(2)`
`impliesx=((alpha+beta)(alphabeta)pmsqrt((alpha+beta)^(2)(alphabeta)^(2)-4(alphabeta)^(3)))/(2)`
`impliesx=((alpha+beta)(alphabeta)pmalphabetasqrt((alpha+beta)^(2)-4alphabeta))/(2)`
`impliesx=alphabeta[((alpha+beta)pmsqrt((alpha-beta)^(2)))/(2)]`
`impliesx=alphabeta[((alpha+beta)pmsqrt((alpha-beta)))/(2)]`
`impliesx=alphabeta[(alpha+beta+alpha-beta)/(2),(alpha+beta-alpha+beta)/(2)]`
`impliesx=alpha beta[(2alpha)/(2),(2 beta)/(2)]`
`impliesx=alpha^(2)beta, alpha beta^(2)` which is the required answer.