Fun With Calculus-3

$\displaystyle \lim_{x \to \infty} \sqrt[3]{(x+a)(x+b)(x+c)} - x$
$\displaystyle \lim_{x \to \infty} \sqrt[3]{x^{3} +(a+b+c)x^{2} + (ab+bc+ca)x + abc} - x$
$\displaystyle \lim_{x \to \infty} x\sqrt[3]{1+\dfrac{a+b+c}{x} + \dfrac{ab+bc+ca}{x^{2}} + \dfrac{abc}{x^{3}}} - x$
Using the binomial expansion for any index
$(1+y)^{n} = 1 + ny + \dfrac{n(n-1)}{2!}y^{2} + \ldots$ when $|y| < 1$

$\displaystyle \lim_{x \to \infty} x\left( 1+\dfrac{1}{3}\left(\dfrac{a+b+c}{x} + \dfrac{ab+bc+ca}{x^{2}} + \dfrac{abc}{x^{3}}\right) - \dfrac{1}{9}\left(\dfrac{a+b+c}{x} + \dfrac{ab+bc+ca}{x^{2}} + \dfrac{abc}{x^{3}}\right)^{2} + \ldots - 1 \right)$
$\displaystyle \lim_{x \to \infty} \dfrac{a+b+c}{3} + x\left( \dfrac{1}{x^{2}} \text{(more terms)}\right)$
All except the first term tend to 0 and limit is equal to,
$\dfrac{a+b+c}{3}$
$x + y + z + d = 1 + 1 + 1 + 3 = 6$