One no. is 18 less than the other no. The HCF of these two no's is 6. And the LCM is 168. The sum of the squares of these two no. is .

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Solution

Let one number is $X$ and other one is $(X + 18)$
$H C F = 6$
$L C M = 168$
First number * second number = HCF \times LCM of the numbers
$X \times (X + 18) = 168 \times 6$
$X^{2} + 18 X - 1008 = 0$
$X^{2} + 42 X - 24 X - 1008 = 0$
$X = - 42$ or $24$
We take positive one.
So, required numbers are $24$ and $44$
So sum of the square = $24 \times 24 + 42 \times 42 = 576 + 1764 = 2340$ $

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