Algebra

\(\frac{rs}{t^2} + \frac{rt}{s^2} + \frac{st}{r^2} \)

Simplify as

rs (rs)^2  + rt (rt)^2  + st (st)^2

________________________       =

            (rst)^2

(rs)^3  + (rt)^3   +  (st)^3

_____________________

           (rst)^2

By Viete

r + s + t =  4

(rs + rt + st)  = -7

(rst)  =  -12

And

(rs + rt + st)^3  =  (-7)^3 = -343 

( rs + rt + st)^3 =

[ (rs)^3 + (rt)^3  + (st)^3 ]  +   3rst [ r^2s + r^2t] + 3rst [ s^2r + s^2t] + 3rst [ t^2r + t^2s] + 6(rst)^2  = 

[ (rs)^3 + (rt)^3 + (st)^3 ]  +   3(-12) [ r ( rs + rt)] + 3(-12) [ s (rs + st)] + 3(-12) [ t(rt + st] + 6(-12)^2  =

[ (rs)^3 + (rt)^3 + (st)^3 ]  -  36 [ r(rs + rt) + s(rs + st)  + t(rt + st) ] + 864

Note: [ rs + rt + st] = -7

So 

rs + rt = -7 - st

rs + st  = -7 - rt

rt + st =  -7 - rs

So

 [(rs)^3 + (rt)^3 + (st)^3 ] -  36 ( r [ -7 - st]  + s [-7 - rt] + t [ -7 -rs] )  + 864  =   -343

[ (rs)^3 + (rt)^3 + (st)^3 ]  - 36 ( -7 [ r + s + t ] - 3 (rst) ]  =   -1207

[ (rs)^3 + (rt)^3 + (st)^3 ]  - 36 [ -7(4) - 3(-12) ]  =  -1207

[ (rs)^3 + (rt)^3 + (st)^3 ] - 36 [ -28 + 36 ]  =  -1207

[ (rs)^3 + (rt)^3 + (st)^3 ]  - 36(8)   =-1207

[ (rs)^3 + (rt)^3 + (st)^3] =  - 288 = -1207

[ (rs)^3 + (rt)^3 + (st)^3]  = -919

So

rs       +     rt      +     st

__           ___          ___  =

r^2           s^2          r^2

[ rs (rs)^2  + rt(rt)^2 + st(st)^2]

________________________   =

       (rst)^2

[ (rs)^3 + (rt)^3 + (st)^3]

___________________  =

       (rst)^2

-919 

____

 144

We are given the cubic equation:

x3−4x2−7x+12=0x^3 - 4x^2 - 7x + 12 = 0

with solutions rr, ss, and tt. We are tasked with finding the value of the expression:

rst2+rts2+str2\frac{rs}{t^2} + \frac{rt}{s^2} + \frac{st}{r^2}

Step 1: Apply Vieta’s Relations

From Vieta's formulas for the cubic equation x3−4x2−7x+12=0x^3 - 4x^2 - 7x + 12 = 0, the coefficients of the polynomial provide the following relationships between the roots rr, ss, and tt:

r+s+t=4(sum of the roots)r + s + t = 4 \quad \text{(sum of the roots)} rs+rt+st=−7(sum of the products of the roots taken two at a time)rs + rt + st = -7 \quad \text{(sum of the products of the roots taken two at a time)} rst=−12(product of the roots)rst = -12 \quad \text{(product of the roots)}

Step 2: Simplify the Expression

We are asked to compute:

rst2+rts2+str2\frac{rs}{t^2} + \frac{rt}{s^2} + \frac{st}{r^2}

To simplify, let’s first combine the terms over a common denominator. The common denominator of t2t^2, s2s^2, and r2r^2 is r2s2t2r^2 s^2 t^2. Thus, the expression becomes:

rst2+rts2+str2=r3s3+r3t3+s3t3r2s2t2\frac{rs}{t^2} + \frac{rt}{s^2} + \frac{st}{r^2} = \frac{r^3 s^3 + r^3 t^3 + s^3 t^3}{r^2 s^2 t^2}

However, we do not need to expand this expression. Instead, we can use a known identity for symmetric sums. Specifically, it is known that:

rst2+rts2+str2=(r+s+t)(rs+rt+st)rst\frac{rs}{t^2} + \frac{rt}{s^2} + \frac{st}{r^2} = \frac{(r+s+t)(rs+rt+st)}{rst}

Step 3: Substitute the Known Values

From Vieta’s relations, we know:

r+s+t=4r + s + t = 4

rs+rt+st=−7rs + rt + st = -7

rst=−12rst = -12

Substituting these values into the identity:

(r+s+t)(rs+rt+st)rst=4×(−7)−12\frac{(r+s+t)(rs+rt+st)}{rst} = \frac{4 \times (-7)}{-12}

Simplifying the expression:

4×(−7)−12=−28−12=2812=73\frac{4 \times (-7)}{-12} = \frac{-28}{-12} = \frac{28}{12} = \frac{7}{3}

Thus, the value of the expression is:

73\boxed{\frac{7}{3}}