a plus b plus c whole square formula
The formula of (a+b+c)2 is utilized to determine the result the of sum of squares involving three numbers without eventually evaluating the squares.
The formula of a plus b plus c whole square is a major algebraic identity. To get the expanded form of (a+b+c)2 formula all you need to do is multiply (a+b+c) twice to determine the whole square of a + b + c.
What is a plus b plus c whole square formula?
We just came to know that if we multiply the expression (a+b+c) twice we can get the (a+b+c)2 formula quite easily.
(a+b+c)2 = (a+b+c) . (a+b+c)
or, (a+b+c)2 = a2 + ac + ab + ba + bc + b2 + ca + c2 + cb
Rearranging the right hand side of this equation we get:
(a+b+c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
or, (a+b+c)2 = a2 + b2 + c2 +2(ab +bc +ac)
Therefore the standard formula of a plus b plus c whole square is:
(a+b+c)2 = a2 + b2 + c2 +2(ab +bc +ac)
1. Find the exact value of a2 + b2 + c2 when a + c + b = – 3.
Also, 1/b + 1/a + 1/c = 3 & abc = 4.
Solution: In this problem we have got three equations:
a + c + b = – 3 … (i)
1/b + 1/a + 1/c = 3 … (ii)
abc = 4 … (iii)
If we multiply equation (ii) with equation (iii)
abc . (1/b + 1/a + 1/c) = 4 x 3
We will have (ac + bc + ab) = 12 … (iv)
At this stage use the a plus b plus c whole square formula.
(a+b+c)2 = a2 + b2 + c2 +2(ab +bc +ca)
Putting the values from equation (i) and (iv)
(-3)2 = a2 + b2 + c2 + 2 x 12
Therefore, a2 + b2 + c2 = 9 – 24 = -15
2. Solve (4x + 5y + 3z)2
Solution: The expression (4x + 5y + 3z)2 resembles the identity of (a+b+c)2
Here, a = 4x, b = 5y and c = 3z
Now put the values replacing a, b and c in the a plus b plus c whole square formula.
(4x + 5y + 3z)2 = (4x)2 + (5y)2 + (3z)2 + 2.4x.5y + 2.5y.3z + 2.3z.4x
or, (4x + 5y + 3z)2 = 16x2 + 25y2 + 9z2 + 40xy + 30yz + 24xz