Question 1044562
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Please help me solve this question in a understandable way for grade 10 curriculum!

Given that 3a - 1/(2a) = 7, determine the value of 27a^3 - 1/(8a^3) without solving for a.
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<pre>
You are given

{{{3a - 1/(2a)}}} = {{{7}}}.       (1)

Square both sides. You will get

{{{9a^2 - 2*3*(1/2)}}} + {{{1/(4a^2)}}} = {{{49}}}.

Or

{{{9a^2 - 3}}} + {{{1/(4a^2)}}} = {{{49}}}.

Add {{{4}}}{{{1/2}}} to both sides. You will get

{{{9a^2 + 3/2}}} + {{{1/(4a^2)}}} = {{{49}}} + {{{4}}}{{{1/2}}} = {{{53}}}{{{1/2}}}.

Notice this one more time: 

{{{9a^2 + 3/2}}} + {{{1/(4a^2)}}} = 53.5.     (2)

Now

  {{{27a^3}}} - {{{1/(8a^3)}}} =      (apply this identity: {{{x^3-y^3}}} = {{{(x-y)*(x^2 + xy + y^2)}}} !)

= {{{(3a - 1/(2a))}}}.({{{9a^2 + 3/2}}} + {{{1/(2a)^2}}}) =   (now recall that the first parentheses is 7 (given !), while the second is 53.5 due to (2) !)

= 7*53.5.

Please calculate the last product on your own.

Is it what you want/need ?
</pre>

Done. 


Why/how I was so smart?


Because I knew the content of the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/HOW-TO-evaluate-expressions-involving-x%2Binv%28x%29-x2%2Binv%28x2%29-and-x%5E3%2Binv%28x%5E3%29.lesson>HOW TO evaluate expressions involving &nbsp;{{{(x + 1/x)}}}, &nbsp;{{{(x^2+1/x^2)}}} &nbsp;and &nbsp;{{{(x^3+1/x^3)}}}</A>

in this site, so, I was prepared in advance.


Now, after reading my post and this lesson, you will be prepared too.


By the way, you have free of charge online textbook in "Algebra-I" in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.