if a(b+c)=119, b(c+a)=128, c(a+b)=135, then find the value of abc
Answers
Answer:
ab=56
bc=72
ca=63
(abc) =√ 56×72×63
abc =504
Given:
a (b + c) = 119---------(i),
b (c + a) = 128--------(ii), and
c (a + b) = 135--------(iii).
To Find:
The value of abc.
Solution:
In equation (i),
ab + ac = 119 -------(iv).
In equation (ii),
bc + ab = 128 ---------(v).
In equation (iii),
ac + bc = 135 ---------(vi).
Adding equation (i), (ii), and (iii),
ab + ac + bc + ab + ac + bc = 119 + 128 + 135.
2ab + 2bc + 2ac = 382.
ab + bc + ac = 191.
We know that, ab + bc = 128.
So,
128 + ac = 191.
ac = 63.
Putting value of ac in equation (iv),
ab + 63 = 119.
ab = 119 - 63.
ab = 56.
Now, Putting value of ab in equation (v),
bc + 56 = 128.
bc = 72.
Now, multiplying the ab, bc, and ca,
ab × bc × ca = 63 ×56 × 72.
a²b²c² = 254016.
abc = √254016.
abc = 504.
Hence, The value of abc is 504.