$\newcommand{\d}{\mathbin{\triangle}}$How do you solve this question; $$1 \mathrel\triangle 0 = 3$$ $$1 \d 2 = 12$$ $$0 \d 2 = 5$$ $$1 \d 5 = 2 \d 3 = 96$$ Find $$2 \d 4 = \text{?}$$ Here it is the answer to the equation(please keep in mind that i need the method);
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Let us write $f(x,y)=x \mathbin\triangle y$. Then by interpolation there is a polynomial $f(x,y)$ satisfying the conditions $$f(1,0)=3$$ $$f(1,2)=12$$ $$f(0,2)=5$$ $$f(1,5)=96$$ $$f(2,3)=96$$ for example
$$ f(x,y)=\frac{1}{110}(708x^3 + 621x^2y - 295xy^2 + 116y^3 - 378). $$ Then we have $$f(2,4)=\frac{6603}{55}.$$ This is not the only possibility, of course.