Let s be the sister's age. We know that each twin is 12 years younger than the sister, or s - 12. The product of their ages would then be
(s - 12)(s - 12)(s). The sum of their ages would be s + (s - 12) + (s - 12). Since the product of their ages is 1244 more than the sum, we have the equation
(s - 12)(s - 12)(s) = s + (s - 12) + (s - 12) + 1244
Multiplying the first two factors on the left (First, Outer, Inner, Last) we have
(s² - 12s - 12s + 144)(s) = s + (s - 12) + (s - 12) + 1244
Combining like terms on both sides we have:
(s² - 24s + 144)(s) = 3s + 1220
Using the distributive property on the left we have
s³ - 24s² + 144s = 3s + 1220
We don't want a variable on both sides, so subtract 3s:
s³ - 24s² + 144s - 3s = 1220
Subtract 1220 from both sides to get our cubic equation to solve:
s³ - 24s² + 144s - 3s - 1220 = 0
Combine like terms:
s³ - 24s² + 141s - 1220 = 0
We want to consider the factors of 1220. After using a factor tree, we get the factors are 2*2*5*61. This gives us possible ages for the sister of 2, 4 (2*2), 4, 10 (2*5), 20 (2*2*5), 61, or older. We can eliminate 2, 4, and 5, since the twins are born and they are 12 years younger than her. This leaves 20, 61 or the older ages (which are improbable; the next age from 61 would be 61*2 = 122).
We will use synthetic division to try 20. Set up the problem with the coefficients of our equation (first image). We bring the 1 down. Now we multiply this 1 by the 20 we're dividing by and put it under the second coefficient (second image). We add these two together now (third image). Continue by multiplying again by 20, then adding, etc. (remaining images; I can't add more than 5, but you keep the pattern going). Since we get 0 at the end, that means that 20 did indeed divide into the problem and is the sister's age. Since the twins are 12 years younger, they are 20-12 = 8.