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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