Question

If a = 24,b = 26,c = 28, then the value of  \[{a^2} + {b^2} + {c^2} - ab - bc - ac\] will be

A. \[0\]

B. \[4\]

C. \[8\]

D. \[12\]

Answer 1

Hint: Here we use the formula for square of addition of three numbers \[a,b,c\]\[{\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right)\]

Solve this question by adding and subtracting \[2(ab + bc + ca)\] to the equation because there already exists \[ - ab - bc - ac\] which will add up to the part that is being subtracted and the other part will help in the formation of the formula. We add and subtract this value so as to suit a formula which can make the calculations easier, otherwise, calculations are very long.

* Here we take one number as the centre and then add or subtract numbers from that particular number to write them in the form of one number. This helps us to avoid calculations with three variables, which will only make our calculations more difficult.

Say we have three numbers \[4,5,6\]. We take \[x = 5\], then we can write \[4 = x - 1\] and \[6 = x + 1\]

So when doing any calculation with three numbers, we will have only one variable that is \[x\].

Complete step by step solution:

Let us take \[b = x\] then \[a = x - 2\]and \[c = x + 2\]

Then, \[x = 26,x - 2 = 24,x + 2 = 28\]

Given, \[a = 24,b = 26,c = 28\],

 Add and subtract \[2(ab + bc + ca)\] to \[{a^2} + {b^2} + {c^2} - ab - bc - ac\] .

\[ = {a^2} + {b^2} + {c^2} - ab - bc - ac + 2\left( {ab + bc + ac} \right) - 2\left( {ab + bc + ac} \right)\]

\[ = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right) - 3\left( {ab + bc + ac} \right)\]

\[ = {\left( {a + b + c} \right)^2} - 3\left( {ab + bc + ac} \right)\] …(1) \[\{ \because {(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2(ab + bc + ca)\} \]

Substitute \[b = x\],\[a = x - 2\] and \[c = x + 2\] in equation (1) .

\[{\left( {x - 2 + x + x + 2} \right)^2} - 3\left[ {\left( {x - 2} \right)x + x\left( {x + 2} \right) + \left( {x - 2} \right)\left( {x + 2} \right)} \right]\]

\[ = {\left( {3x} \right)^2} - 3\left[ {{x^2} - 2x + {x^2} + 2x + {x^2} - 4} \right]\] \[\left\{ {\because \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right\}\]

\[ = {\left( {3x} \right)^2} - 3\left[ {({x^2} + {x^2} + {x^2}) + (2x - 2x) - 4} \right]\]

\[ = {\left( {3x} \right)^2} - 3\left[ {3{x^2} - 4} \right]\]

Simplify the parentheses.

\[ = 9{x^2} - 3 \times (3{x^2}) - 3 \times ( - 4)\]

\[ = 9{x^2} - 9{x^2} + 12\]

\[ = 12\]

Therefore, Option (D) is correct.

Note:

In these types of questions where we can see the formulas in a twisted way, always add or subtract some values to make a formula which can be used there.

Alternative method:

This method involves lots of calculations and students who have a good hand on squares of numbers can opt for this method. It is easier than the other method as there are no assumptions, just substitution of numbers into the formula.

Students should do calculations very carefully.

Substitute the values \[a = 24,b = 26,c = 28\] directly in the equation \[{a^2} + {b^2} + {c^2} - ab - bc - ac\]

\[{a^2} + {b^2} + {c^2} - ab - bc - ac = {(24)^2} + {(26)^2} + {(28)^2} - (24 \times 26) - (26 \times 28) - (28 \times 24)\]

\[ = 576 + 676 + 784 - 624 - 728 - 672\]

\[ = 2036 - 2024 = 12\]