Question

Please help guys question number 15

Answer 1

Hope u like my process
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Formula to be used
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$= > {a}^{3} - {b}^{3} = (a - b)( {a}^{2} +a b + {b}^{2} )$
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Let one of the positive number be x

Other number = (x-4)
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By problem
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$= > {a}^{3} - {b}^{3} = 316 \\ \\ = > {(x )}^{3} - {(x - 4)}^{3} = 316 \\ \\ = > (x - (x - 4))( {x}^{2} + x(x - 4) + {(x - 4)}^{2} ) = 316 \\ \\ = > 4( {x}^{2} + {x}^{2} - 4x + {x }^{2} - 8x + 16) = 316 \\ \\ = > (3 {x}^{2} - 12x + 16) = \frac{316}{4} \\ \\ = > 3 {x}^{2} - 12x + 16 - 79 = 0 \\ \\ = > 3 {x}^{2} - 12x - 63 = 0 \\ \\ = > {x}^{2} - 4x - 21 = 0 \\ \\ = > {x}^{2} - 7x + 3x - 21 = 0 \\ \\ = > x(x - 7) + 3(x - 7) = 0 \\ \\ = > (x - 3)(x - 7) = 0 \\ \\ \bf \underline{since \: \: the \: \: number \: \: is \: \: positive} \\ \\ = > x - 7 = 0 \\ \\ = > \bf \: x = \underline{7}$

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So

=> 1st positive number =x= 7

=> 2nd positive number =(x-4) =(7-4)=3

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ɪ) ᴛʜᴇɪʀ ᴩʀᴏᴅᴜᴄᴛ = 7×3=21

ɪɪ) ꜱᴜᴍ ᴏꜰ ᴛʜᴇɪʀ ꜱqᴜᴀʀᴇꜱ =7²+3²

= 49 + 9 = 58

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Hope this is ur required answer

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Answer 2

Answer :

Let two positive numbers are a  And  b  . 
As given 

a  -  b  =  4         ------ ( 1 )

And,

a^3  - b^3  = 316   ------ ( 2 ) 

i ) Their products 

We know,

( a -  b  )^3  = a^3 - b^3  - 3ab ( a -  b  )                      

Substitute all values from equation 1 and 2

We get,

(  4  )^3 = 316 - 3ab ( 4  )

64 = 316 -   12 ab 

12ab  = 252 

ab   =   21         ----- ( 3 )

So,

Their product will be   =  21  ( Answer  )

ii ) The sum of their squares

We know,

a^3 - b^3   = ( a  - b ) ( a^2  + ab  + b^2 ) 

Substitute all values from equation 1 , 2 and 3

We get,

316  = 4 ( a^2 + 21  + b^2 ) 

( a^2 + 21  + b^2 )  = 79 

 a^2 +  b^2   =   79   -    21 

 a^2 +  b^2  =   58  

So,

The sum of their squares   =    58  (Answer)