<s>[INST] 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.
 [/INST] 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 $a,b,c\space\space\space\space\space\space\blacksquare$
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