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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