Question

Find a,b,c. when abc=144,2(ab+bc+ac)=192 and a^2+b^2+c^2=169

Answer 1

Applying the following identity,
${(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)$
${(a + b + c)}^{2} = 169 + 192 = 361$
$a + b + c = 19 \: - - - - - (1)$
Given abc = 144 ------(2)

Considering (1) & (2) together,
The values of a, b, c could be combination of
2, 8, 9 or 3, 4, 12

Answer 2

a = 3, b = 4  and c = 12

Step-by-step explanation:

We have,

abc = 144, 2(ab + bc + ac) = 192 and $a^2+b^2+c^2=169$

To find, the values of a, b and c = ?

We know that,

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

Put abc = 144, 2(ab + bc + ac) = 192 and $a^2+b^2+c^2=169$, we get

⇒ $(a+b+c)^{2}=169+192$

⇒ $(a+b+c)^{2}=361$

⇒ $(a+b+c)^{2}=19^2$

⇒ a + b + c = 19

To find the values of a, b and c by putting value method,

a = 3, b = 4  and c = 12

Satiesfied all conditions.

Hence, a = 3, b = 4  and c = 12