Show the conjecture is false by finding a counterexample.
the product (a+b)^2 is equal to a^2+b^2, for a is not equal to 0 and be not equal to 0.
it’s number 15

(a+b)^2=(a+b)*(a+b)

Open the brackets

a*a+a*b+b*a+b*b

Doing the operations We get

(a+b)^2=a^2+2a*b+b^2

If both were equal then`

a^2+2ab+b^2=a^2+b^2

Subtract (a^2-b^2) from both sides.

We are left with 2ab=0

This equation can be true only and only if either a and/or b are equal to 0.

If they are not then the statement can not be true.

Another way of doing this is to put integers instead of variable. But You have to be careful because this way You only prove that statement is not true universally, by the way it can still be true for some other values. So it's better and precise practice to go for the first solution I provided You.

(a) the volume of the cone is 20/3 in3

hence, option a is correct.

but there’s no any figure there!