Expansion of (a ± b ± c)\(^{2}\)

We will discuss here about the expansion of (a ± b ± c)\(^{2}\).

(a + b + c)\(^{2}\) = {a + (b + c)}\(^{2}\) = a\(^{2}\) + 2a(b + c) + (b + c)\(^{2}\)

= a\(^{2}\) + 2ab + 2ac + b\(^{2}\) + 2bc + c\(^{2}\)

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

= sum of squares of a, b, c + 2(sum of the products of a, b, c taking two at a time}.

Therefore, (a – b + c)\(^{2}\) = a\(^{2}\) + b\(^{2}\) + c\(^{2}\) + 2(ac – ab – bc)

Similarly for (a – b – c)\(^{2}\), etc.

Corollaries:

(i) a\(^{2}\) + b\(^{2}\) + c\(^{2}\) = (a + b + c)\(^{2}\) – 2(ab + bc + ca)

(ii) ab + bc + ca = \(\frac{1}{2}\){(a + b + c)\(^{2}\) – (a\(^{2}\) + b\(^{2}\) + c\(^{2}\))}


Solved Examples on Expansion of (a ± b ± c)\(^{2}\)

1. Expand (2x + y +3z)^2

Solution:

(2x + y +3z)\(^{2}\)

= (2x)\(^{2}\) + y\(^{2}\) + (3z)\(^{2}\) + 2{2x ∙ y + y ∙ 3z + 3z ∙ 2x}

= 4x\(^{2}\) + y\(^{2}\) + 9z\(^{2}\) + 4xy + 6yz + 12zx.

 

2. Expand (a - b - c)\(^{2}\)

Solution:

(a - b - c)\(^{2}\)

= a\(^{2}\) + (-b)\(^{2}\) + (-c)\(^{2}\) + 2{a ∙ (-b) + (-b) ∙ (-c) + (-c) ∙ a}

= a\(^{2}\) + b\(^{2}\) + c\(^{2}\) - 2ab + 2bc - 2ca.

 

3. Expand (m - \(\frac{1}{2x}\) + m\(^{2}\))\(^{2}\)

Solution:

(m - \(\frac{1}{2x}\) + m\(^{2}\))\(^{2}\)

m\(^{2}\) + (-\(\frac{1}{2m}\))\(^{2}\) + (m\(^{2}\))\(^{2}\) + 2{m ∙ (-\(\frac{1}{2m}\)) + (-\(\frac{1}{2m}\)) ∙ m\(^{2}\)  + m\(^{2}\)  ∙ m}

= m\(^{2}\) + \(\frac{1}{4m^{2}}\)+ m\(^{4}\) + 2{-\(\frac{1}{2}\) - \(\frac{1}{2}\)m + m\(^{3}\)}

= m\(^{2}\) + \(\frac{1}{4m^{2}}\)+ m\(^{4}\) - 1 - m + 2m\(^{3}\).


4. If p + q + r = 8 and pq + qr + rp = 18, find the value of p\(^{2}\) + q\(^{2}\) + r\(^{2}\).

Solution:

We know that p\(^{2}\) + q\(^{2}\) + r\(^{2}\) = (p + q + r)\(^{2}\) - 2(pq + qr + rp).

Therefore, p\(^{2}\) + q\(^{2}\) + r\(^{2}\)

= 8\(^{2}\) - 2 × 18

= 64 – 36

= 28.


5. If x – y – z = 5 and x\(^{2}\) + y\(^{2}\) + z\(^{2}\) = 29, find the value of xy – yz – zx.

Solution:

We know that ab + bc + ca = \(\frac{1}{2}\)[(a + b + c)\(^{2}\) – (a\(^{2}\) + b\(^{2}\) + c\(^{2}\))].

Therefore, xy + y(-z) + (-z)x = \(\frac{1}{2}\)[(x + y - z)\(^{2}\) – (x\(^{2}\) + y\(^{2}\) + (-z)\(^{2}\))]

Or, xy – yz – zx = \(\frac{1}{2}\)[5\(^{2}\) – (x\(^{2}\) + y\(^{2}\) + z\(^{2}\))]

                        = \(\frac{1}{2}\)[25 – 29]

                        = \(\frac{1}{2}\)(-4)

                        = -2.










9th Grade Math

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