Question

What is the length of the edge of a cube if, after

a slice $1$inch thick is cut from one side, the volume remaining is

$294$cubic inches?

Short answer

If The remaining volume of the cube after cutting a one-inch slice from one side is $294$cubic inches then the original side of the cube was$7$inches

Step-by-step solution

Step 1. given data

The remaining volume of the cube after cutting a one-inch slice from one side is$294$cubic inches

Step 2. Formation of equation

consider a side of the cube is x

then volume $V=x^3$

after cutting a slice of one inch from one side

$$\begin{gathered} x^2(x-1)=294 \\ {x}^3-x^2=294 \\ {x}^3-x^2+0x-294=0 \end{gathered}$$

Step 3. possible zeros

the degree of Equation $x^3-x^2+0x-294=0$is $3$so maximum solutions can be $3$

the constant term is $-294$and the leading coefficient is $1$

so potential rational zeros are

$\pm 1,\pm 2,\pm 3,\pm 6,\pm 7,\pm 14,\pm 21,\pm 42,\pm 49,\pm 98,\pm 147$

Step 4. testing potential rational zeros 

Test potential rational zeros $6$and factor the equation by synthetic division

$\frac{6\begin{array}{cccc}1& -1& 0& -294\\ & 6& 30& 180\end{array}}{\begin{array}{cccc}1& 5& 30& -114\end{array}}$

Step 4. testing potential rational zeros 

Test potential rational zeros $7$and factor the equation by synthetic division

$7\begin{array}{cccc}1& -1& 0& -294\end{array}\begin{array}{cccc}& 7& 42& 294\end{array}\begin{array}{cccc}1& 6& 42& 0\end{array}$

so $f\left(7\right)=0$and $7$is a solution of equation

the equation can be factorized as$(x-7)(x^2+6x+42)=0$

discriminant of the depressed quadratic equation $x^2+6x+42=0$is

$$\begin{gathered} b^2-4ac=6^2-4\left(1\right)\left(42\right) \\ =-132 \end{gathered}$$

the discriminant of the quadratic equation is negative so it does not have any real solution

so the real solution of the equation$x^3-x^2-194=0$is$7$and side to the cube was$7$inches