Roots of $(x-\alpha)(x-\beta)(x-\gamma)+d=0$? 
Let $\alpha,\beta,\gamma$ be the roots of $$(x-a)(x-b)(x-c)=d,d\neq 0$$ then the roots of $$(x-\alpha)(x-\beta)(x-\gamma)+d=0?$$

My tries:
Clearly $$(x-a)(x-b)(x-c)=-(x-\alpha)(x-\beta)(x-\gamma)=d\tag{1}$$
As $\alpha,\beta,\gamma$ are roots of equation on left, also if we'll put $x=a,b,c$ gives $0$ on $LHS\implies$ $a,b,c$ are roots of: $$(x-\alpha)(x-\beta)(x-\gamma)=0\\\underbrace{(x-a)(x-b)(x-c)}_{0\ at\ x=a,b,c}+(x-\alpha)(x-\beta)(x-\gamma)=0\rightarrow a,b,c\ \text{satisfy this.}\\(x-\alpha)(x-\beta)(x-\gamma)+d=0\ \rightarrow a,b,c\ \text{satisfy this.}$$.
Hence the result, $a,b,c$ is the answer.
But when we put in $x=\alpha,\beta,\gamma$ in $(1)$, gives $$(x-a)(x-b)(x-c)=0=d$$ but $d\neq 0$.
What I'm mistaking, please help.
 A: As the roots of the polynomial 
$$(x-a)(x-b)(x-c)-d$$ are $\alpha,\beta$ and $\gamma$, we can conclude that, by the fundamental theorem of algebra, and as the coefficient of $x^3$ is $1$,
$$(x-a)(x-b)(x-c)-d=(x-\alpha)(x-\beta)(x-\gamma)$$
Then,
$$(x-\alpha)(x-\beta)(x-\gamma)+d=0$$
$$\iff (x-a)(x-b)(x-c)-d+d=0$$
$$\iff (x-a)(x-b)(x-c)=0$$
A: From $(x-a)(x-b)(x-c)=d$ we have $$a+b+c=\alpha+\beta+\gamma\\ab+bc+ca=\alpha\beta+\beta\gamma+\gamma\alpha\\abc+d=\alpha\beta\gamma$$
Now $(x-\alpha)(x-\beta)(x-\gamma)+d=x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma+d\\=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc-d+d\\=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=(x-a)(x-b)(x-c)$
Hence roots are $a,b,c\space\space\space\space\space\space\blacksquare$
A: Note that $\alpha$, $\beta$ and $\gamma$ are roots of the equation, $(x-a)(x-b)(x-c)=d$.
Hence it follows that,
$$
(\alpha-a)(\beta-b)(\gamma-c)=d
$$
or, 
$$
(a-\alpha)(b-\beta)(c-\gamma)=-d
$$
Now observe that this clearly shows that $a$, $b$ and $c$ are roots of the equation $(x-\alpha)(x-\beta)(x-\gamma)=-d$

QED
