In binomial expansion of (ax² + bx + c)(1 – 2x)²⁶, the coefficients of x, x² and x³ are –56, 0 and 0 respectively, then (a + b + c) is equal to

To find the value of $a+b+c$, we need to determine the coefficients of $x$, $x^2$, and $x^3$ in the expansion of the given expression:
$P\left(x\right)=(a{x}^2+bx+c){(1-2x)}^{26}$

First, let us expand the binomial term ${(1-2x)}^{26}$ using the binomial theorem:
${(1-2x)}^{26}=\left(\genfrac{}{}{0ex}{}{26}{0}\right)-\left(\genfrac{}{}{0ex}{}{26}{1}\right)\left(2x\right)+\left(\genfrac{}{}{0ex}{}{26}{2}\right){\left(2x\right)}^2-\left(\genfrac{}{}{0ex}{}{26}{3}\right){\left(2x\right)}^3+...$

Calculating the binomial coefficients:
- First term: $1$
- Second term: $-26·2x=-52x$
- Third term: $\frac{26·25}{2}·4x^2=325·4x^2=1300x^2$
- Fourth term: $-\frac{26·25·24}{6}·8x^3=-2600·8x^3=-20800x^3$

Therefore, we can write:
${(1-2x)}^{26}=1-52x+1300x^2-20800x^3+...$

Now, we substitute this series back into the expression:
$P\left(x\right)=(a{x}^2+bx+c)(1-52x+1300x^2-20800x^3+...)$

Let us write down the coefficients of $x$, $x^2$, and $x^3$ by multiplying the polynomial terms:

1. Coefficient of $x$:
$\text{Coefficient of}x=b\left(1\right)+c(-52)=b-52c$
Given that the coefficient of $x$ is $-56$:
$b-52c=-56 \text{(Equation 1)}$

2. Coefficient of $x^2$:
$\text{Coefficient of}{x}^2=a\left(1\right)+b(-52)+c\left(1300\right)=a-52b+1300c$
Given that the coefficient of $x^2$ is $0$:
$a-52b+1300c=0 \text{(Equation 2)}$

3. Coefficient of $x^3$:
$\text{Coefficient of}{x}^3=a(-52)+b\left(1300\right)+c(-20800)=-52a+1300b-20800c$
Given that the coefficient of $x^3$ is $0$:
$-52a+1300b-20800c=0$
Dividing the entire equation by $-52$:
$a-25b+400c=0 \text{(Equation 3)}$

Now, we solve the system of equations. Subtract Equation 3 from Equation 2:
$(a-52b+1300c)-(a-25b+400c)=0-0$
$-27b+900c=0$
$27b=900c$
$b=\frac{900}{27}c=\frac{100}{3}c \text{(Equation 4)}$

Substitute Equation 4 into Equation 1:
$\frac{100}{3}c-52c=-56$
Multiply the entire equation by $3$ to eliminate the denominator:
$100c-156c=-168$
$-56c=-168$
$c=3$

Substitute $c=3$ back into Equation 4:
$b=\frac{100}{3}\left(3\right)=100$

Substitute $b=100$ and $c=3$ into Equation 3 to find $a$:
$a-25\left(100\right)+400\left(3\right)=0$
$a-2500+1200=0$
$a-1300=0$
$a=1300$

Finally, we compute $a+b+c$:
$a+b+c=1300+100+3=1403$

Thus, the value of $(a+b+c)$ is equal to 1403.