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# The sum of three numbers in A.P. is 12, and the sum of their cubes is 408; find them.  Verified
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Hint: Assume the numbers as (a – d), a, (a + d) and apply the conditions to solve to get the values of ‘a’ and ‘d’ and after that put that values in assumed numbers to get the answer.

As we have given that the three numbers are in A.P. and therefore we will assume (a-d), a, (a+d) as the three numbers in A.P. with ‘a’ as the first term and‘d’ is the common difference.
As the sum of three numbers is 12, therefore we can write,
(a - d) + a + (a + d) = 12
$\Rightarrow$a - d + a + a + d = 12
$\Rightarrow$a + a + a = 12
$\Rightarrow$ 3a = 12
$\Rightarrow a=\dfrac{12}{3}$
$\Rightarrow$ a = 4 …………………………………. (2)
Now, as per the second condition given in the problem we can write,
$\therefore {{\left( a-d \right)}^{3}}+{{a}^{3}}+{{\left( a+d \right)}^{3}}=408$
If we put the value equation (2) in the middle term of the above equation we will get,
$\Rightarrow {{\left( a-d \right)}^{3}}+{{4}^{3}}+{{\left( a+d \right)}^{3}}=408$
$\Rightarrow {{\left( a-d \right)}^{3}}+64+{{\left( a+d \right)}^{3}}=408$
$\Rightarrow {{\left( a-d \right)}^{3}}+{{\left( a+d \right)}^{3}}=408-64$
$\Rightarrow {{\left( a-d \right)}^{3}}+{{\left( a+d \right)}^{3}}=344$ ………………………………… (3)
Now to proceed further in the solution we should know the formula given below,
Formula:
$\left( {{x}^{3}}+{{y}^{3}} \right)=\left( x+y \right)\times \left( {{x}^{2}}+{{y}^{2}}-xy \right)$
By using above formula we can write equation (3) as follows,
$\therefore \left[ \left( a-d \right)+\left( a+d \right) \right]\times \left[ {{\left( a-d \right)}^{2}}+{{\left( a+d \right)}^{2}}-\left( a-d \right)\times \left( a+d \right) \right]=344$
$\therefore \left[ a-d+a+d \right]\times \left[ {{\left( a-d \right)}^{2}}+{{\left( a+d \right)}^{2}}-\left( a-d \right)\times \left( a+d \right) \right]=344$
$\therefore \left[ a+a \right]\times \left[ {{\left( a-d \right)}^{2}}+{{\left( a+d \right)}^{2}}-\left( {{a}^{2}}+ad-ad-{{d}^{2}} \right) \right]=344$
$\therefore 2a\times \left[ {{\left( a-d \right)}^{2}}+{{\left( a+d \right)}^{2}}-\left( {{a}^{2}}-{{d}^{2}} \right) \right]=344$
$\therefore 2a\times \left[ {{\left( a-d \right)}^{2}}+{{\left( a+d \right)}^{2}}-{{a}^{2}}+{{d}^{2}} \right]=344$
Now to proceed further we should know the formulae given below,
Formulae:
${{\left( x+y \right)}^{2}}=\left( {{x}^{2}}+2xy+{{y}^{2}} \right)$ And ${{\left( x+y \right)}^{2}}=\left( {{x}^{2}}-2xy+{{y}^{2}} \right)$
By using the above formulae we can write the above equation as,
$\therefore 2a\times \left[ \left( {{a}^{2}}-2ad+{{d}^{2}} \right)+\left( {{a}^{2}}+2ad+{{d}^{2}} \right)-{{a}^{2}}+{{d}^{2}} \right]=344$
By opening the brackets we will get,
$\Rightarrow 2a\times \left[ {{a}^{2}}-2ad+{{d}^{2}}+{{a}^{2}}+2ad+{{d}^{2}}-{{a}^{2}}+{{d}^{2}} \right]=344$
$\Rightarrow 2a\times \left[ {{d}^{2}}+{{a}^{2}}+{{d}^{2}}+{{d}^{2}} \right]=344$
$\Rightarrow 2a\times \left[ 3{{d}^{2}}+{{a}^{2}} \right]=344$
By substituting the value of equation (2) in the above equation we will get,
$\Rightarrow 2\times 4\times \left[ 3{{d}^{2}}+{{4}^{2}} \right]=344$
$\Rightarrow 8\times \left[ 3{{d}^{2}}+{{4}^{2}} \right]=344$
${{\left( x+y \right)}^{3}}and{{\left( x-y \right)}^{3}}$
$\Rightarrow 3{{d}^{2}}+16=43$
$\Rightarrow 3{{d}^{2}}=43-16$
$\Rightarrow 3{{d}^{2}}=27$
$\Rightarrow {{d}^{2}}=\dfrac{27}{3}$
$\Rightarrow {{d}^{2}}=9$
By taking square roots on both sides of the equation we will get,
$\therefore d=\pm 3$
Therefore d = 3 OR d = -3 ……………………………………. (4)
Now we will rewrite the three numbers below,
(a – d), a, (a + d)
If we put the values of equation (2) and equation (4) in above equation as shown below,
a = 4 and d = 3
Therefore numbers will become,
(4 – 3), 4, (4+3)
Therefore the numbers are,
1, 4, 7.
Now,
a = 4 and d = - 3
Therefore numbers will become,
[4 – (-3)], 4, [4+(-3)]
(4 + 3), 4, (4 – 3)
Therefore the numbers are,
7, 4, 1.
Therefore the three numbers are 1, 4, 7 or 7, 4, 1.

Note: Assume the standard numbers given by (a – d), a, (a + d) to make the calculations easier. Also in the step ${{\left( a-d \right)}^{3}}+{{\left( a+d \right)}^{3}}=344$ you can also use the formulae of ${{\left( x+y \right)}^{3}}and{{\left( x-y \right)}^{3}}$