If $3a -2b +5c = 5$ and $6ab +10bc -15ac =14$, then find the value of $27a^3 +125b^3+ 90abc-8b^3$

Given :

The given expressions are $3a -2b +5c = 5$ and $6ab +10bc -15ac =14$

To find :

We have to find the value of   $27a^3 +125b^3+ 90abc-8b^3$

Solution :

We know that,

$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

$3a-2b+5c = 5$

Squaring on both sides, we get,

$(3a-2b+5c)^2= 5^2$

$(3a)^2+(-2b)^2+(5c)^2+2(3a)(-2b)+2(-2b)(5c)+2(5c)(3a) = 25$

$9a^2+4b^2+25c^2-12ab-20bc+30ac = 25$

$9a^2+4b^2+25c^2-2(6ab+10bc-15ac) = 25$

$9a^2+4b^2+25c^2-2(14) = 25$

$9a^2+4b^2+25c^2 = 25+28$

$9a^2+4b^2+25c^2 = 53$ ------(1)

We know that,

$a^3 + b^3+ c^3 -3abc = (a+b+c) (a^2+b^2+c^2-ab-bc-ca) $

Given,

$27a^3 +125c^3+ 90abc-8b^3$ 

It can be written as,

$27a^3 +125c^3+ 90abc-8b^3 = (3a)^3 +(-2b)^3+(5c)^3-3(3a)(-2b)(5c)$

Therefore,

$27a^3+125c^3+ 90abc-8b^3= (3a)^3+(-2b)^3+(5c)^3-3(3a)(-2b)(5c)$

                                                 $=(3a-2b+5c)((3a)^2+(-2b)^2+(5c)^2-(3a)(-2b)-(-2b)(5c)-(5c)(3a)$

                                               $ =(5)(9a^2+4b^2+25c^2 +6ab+10bc-15ac) $  

                                               $ =(5)(53+14)$      (From equation (1))

                                               = 5(67)

                                               = 335.

The value of  $27a^3 +125b^3+ 90abc-8b^3$ is 335

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Updated on: 10-Oct-2022

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