Question

Three numbers are such that their product is three times the sum of their cubes. What is their sum if the sum of their squares and the product taken two at a time is both 15.6? ( A ) 4.6 ( B ) 5.6 ( C ) 0 ( D ) 6.5

Answer 1

Their sum is 6.84105.

Step-by-step explanation:

We assume the three numbers are a, b and c.

According to the question, their product is 3 times the sum of their cubes.

So, $a \times b \times c = 3 (a^3 + b^3 + c^3)$

$abc = 3 (a^3 + b^3 + c^3)$   .....(1)

Also, the sum of their squares and the product taken two at a time is both 15.6.

So,$(a^2 + b^2 + c^2) = 15.6$        .......(2)

ab + bc + ca = 15.6                       ...........(3)

We know that;

• $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc +ca)$

Put the values of expressions in above formula, we get as;

• $(a + b + c)^2 = 15.6 + (2 \times 15.6)$   ........From (2) & (3)

$(a + b + c)^2 = 15.6 + 31.2$

$(a + b + c)^2 = 46.8$

$a + b + c = 6.84105$

Thus, their sum is 6.84105.