Question

$(x-1{)}^4+4(x-1{)}^3+6(x-1{)}^2+4(x-1)+1=$

• A. $x^4$
• B. $x^3$
• C. $x^2$
• D. $1$

Answer 1

The correct option is A $x^4$

Consider the following identity
$(a+1){}^4$
$=\left(\right(a+1\left){}^2\right){}^2$
$=(a^2+2a+1){}^2$
$=a^4+4a^2+1+4a^3+4a+2a^2$
$=a^4+4a^3+6a^2+4a+1$ $...\left(i\right)$
Comparing i with the given question we get
$a=(x-1)$
Therefore
$(x-1){}^4+4(x-1){}^3+6(x-1){}^2+4(x-1)+1$
$=(x-1+1){}^4$ from $\left(i\right)$
$=x^4.$