I got x = 4 as my answer, but are there any other values as it is a high marked question.

Question

The dimensions of a box are 3 cm × 5 cm × 7 cm. The width, length, and height of the box must be increased by the same amount x in order for the box to have a volume of 693 cm cubed. Determine the value of x that will produce a box with a volume of 693 cm3.

Answer 1

(3+x)(5+x)(7+x) = 693

(x^2 +8x +15)(x+7) = 693

x^3 + 15x^2 + 71x + 105 = 693 cubic has 3 solutions, only 1 real given only 1 change in sign

x^3 + 15x^2 + 71x - 588 = 0 use synthetic or long division, dividy by x-4

(x-4)(x^2+19x+147) = 0

there's only 1 real solution. since the discriminant <0 for the 2nd factor

x = 4, or two other imaginary solutions, setting the 2nd factor =0 and solving for x with the quadratic formula

x= -19/2 + or - (1/2)sqr(19^2 - 4(147)) = -9.5 + or - .5sqr-227 = -9.5 + or - about 7.5i

(3+4)(5+4)(7+4) = 7(9)11 = 7(99)= 693 cm^3

that's the only solution unless you allow imaginary dimensions for the box