Answer the following questions based on the evaluation of a sum (without the help of calculator) below:

Evaluate [MATH] \frac{(-\sqrt{6}+\sqrt{7}+\sqrt{8})^4}{4(\sqrt{7}-\sqrt{6})(\sqrt{8}-\sqrt{6})}+\frac{(\sqrt{6}-\sqrt{7}+\sqrt{8})^4}{4(\sqrt{6}-\sqrt{7})(\sqrt{8}-\sqrt{7})}+\frac{(\sqrt{6}+\sqrt{7}-\sqrt{8})^4}{4(\sqrt{6}-\sqrt{8})(\sqrt{7}-\sqrt{8})}[/MATH].

Question 1: Would you manage the fraction one by one by simplifying them and then evaluate the sum from there?

A. Yes.

B. No.

Answer:

More often than not, question like this doesn't encourage us to handle each of the fraction before adding them up. It must be correct that the sum can greatly be simplified so one doesn't has to rely on calculator to add it.

Question 2: Would you rationalize the fractions in order to proceed?

A. Yes.

B. No.

Answer:

If you're less sure about yourself, you could always try to rationalize the fractions to see if it leads you in the right direction by making the expression less complicated and there is much to be simplified as you go:

[MATH] \frac{(-\sqrt{6}+\sqrt{7}+\sqrt{8})^4}{4(\sqrt{7}-\sqrt{6})(\sqrt{8}-\sqrt{6})}+\frac{(\sqrt{6}-\sqrt{7}+\sqrt{8})^4}{4(\sqrt{6}-\sqrt{7})(\sqrt{8}-\sqrt{7})}+\frac{(\sqrt{6}+\sqrt{7}-\sqrt{8})^4}{4(\sqrt{6}-\sqrt{8})(\sqrt{7}-\sqrt{8})}[/MATH]

[MATH]= \frac{(-\sqrt{6}+\sqrt{7}+\sqrt{8})^4}{4(\sqrt{7}-\sqrt{6})(\sqrt{8}-\sqrt{6})}\frac{(\sqrt{7}+\sqrt{6})(\sqrt{8}+\sqrt{6})}{(\sqrt{7}+\sqrt{6})(\sqrt{8}+\sqrt{6})}+\frac{(\sqrt{6}-\sqrt{7}+\sqrt{8})^4}{4(\sqrt{6}-\sqrt{7})(\sqrt{8}-\sqrt{7})}\frac{(\sqrt{6}+\sqrt{7})(\sqrt{8}+\sqrt{7})}{(\sqrt{6}+\sqrt{7})(\sqrt{8}+\sqrt{7})}[/MATH]

[MATH]\,\,\,\,\,\,\,\,+\frac{(\sqrt{6}+\sqrt{7}-\sqrt{8})^4}{4(\sqrt{6}-\sqrt{8})(\sqrt{7}-\sqrt{8})}\frac{(\sqrt{6}+\sqrt{8})(\sqrt{7}+\sqrt{8})}{(\sqrt{6}+\sqrt{8})(\sqrt{7}+\sqrt{8})}[/MATH]

[MATH]= \frac{(-\sqrt{6}+\sqrt{7}+\sqrt{8})^4(\sqrt{7}+\sqrt{6})(\sqrt{8}+\sqrt{6})}{4(1)(2)}+\frac{(\sqrt{6}-\sqrt{7}+\sqrt{8})^4(\sqrt{6}+\sqrt{7})(\sqrt{8}+\sqrt{7})}{4(-1)(1)}[/MATH]

[MATH]\,\,\,\,\,\,\,\,+\frac{(\sqrt{6}+\sqrt{7}-\sqrt{8})^4(\sqrt{6}+\sqrt{8})(\sqrt{7}+\sqrt{8})}{4(-2)(-1)}[/MATH]

[MATH]= \frac{(-\sqrt{6}+\sqrt{7}+\sqrt{8})^4(\sqrt{7}+\sqrt{6})(\sqrt{8}+\sqrt{6})}{8}-\frac{(\sqrt{6}-\sqrt{7}+\sqrt{8})^4(\sqrt{6}+\sqrt{7})(\sqrt{8}+\sqrt{7})}{4}[/MATH]

[MATH]\,\,\,\,\,\,\,\,+\frac{(\sqrt{6}+\sqrt{7}-\sqrt{8})^4(\sqrt{6}+\sqrt{8})(\sqrt{7}+\sqrt{8})}{8}[/MATH]

Stop and see what do we get so far...it's not hard to see there is no common factors between the three, and it would be unwise to continue with what we've started, i.e. to rationalize the denominators of the three fractions.

Question 3: Would you multiply out the 4th power of $(-\sqrt{6}+\sqrt{7}+\sqrt{8})^4,\,(\sqrt{6}-\sqrt{7}+\sqrt{8})^4,\,(\sqrt{6}+\sqrt{7}-\sqrt{8})^4,\,$ to proceed?

A. Yes.

B. No.

Answer:

Multiply out the 4th power of a binomial is already a tedious process, what to mention about multiplying out the 4th power of a trinomial? That is simply dissuading. Plus, we're not asking to evaluate $(-\sqrt{6}+\sqrt{7}+\sqrt{8})^4+(\sqrt{6}-\sqrt{7}+\sqrt{8})^4+(\sqrt{6}+\sqrt{7}-\sqrt{8})^4$, the expression that we're given looks more complicated than this, so multiply out the 4th power in order to proceed is out of the question.

Question 4: You're asked to let $a=-\sqrt{6}+\sqrt{7}+\sqrt{8}),\,b=\sqrt{6}-\sqrt{7}+\sqrt{8},\,c=\sqrt{6}+\sqrt{7}-\sqrt{8}$, what would you do to relate the denominators in terms of only $a,\,b$ and $c$?

A. By adding the equations $a,\,b$ and $c$.

B. By subtracting the equations $a,\,b$ and $c$.

C. By multiplying the equations $a,\,b$ and $c$.

D. By dividing the equations $a,\,b$ and $c$.

Answer:

If

$a=-\sqrt{6}+\sqrt{7}+\sqrt{8}$

$b=\sqrt{6}-\sqrt{7}+\sqrt{8}$

$c=\sqrt{6}+\sqrt{7}-\sqrt{8}$

and the denominators are $4(\sqrt{7}-\sqrt{6})(\sqrt{8}-\sqrt{6}),\,4(\sqrt{6}-\sqrt{7})(\sqrt{8}-\sqrt{7}),\,4(\sqrt{6}-\sqrt{8})(\sqrt{7}-\sqrt{8})$, it's quite obvious that:

$\sqrt{7}-\sqrt{6}=\dfrac{a-b}{2}$

$\sqrt{8}-\sqrt{6}=\dfrac{a-c}{2}$

$\sqrt{8}-\sqrt{7}=\dfrac{b-c}{2}$

Therefore, it's the subtraction between the equations $a,\,b$ and $c$ that related nicely between the denominators and $a,\,b$ and $c$.

[MATH] \frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-c)(b-a)}+\frac{c^4}{(c-a)(c-b)}[/MATH]

Question 5: You're asked to add up the expression above. What would be the best possible form that you would use to evaluate the sum?

A.[MATH]\frac{a^4(b-c)+b^4(a-c)+c^4(a-c-b+c)}{(a-b)(b-c)(a-c)}[/MATH]

B. [MATH]\frac{a^4(b-c)+b^4(a-c)+c^4(a-b)}{(a-b)(b-c)(a-c)}[/MATH]

Answer:

If we've to add up the expression [MATH] \frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-c)(b-a)}+\frac{c^4}{(c-a)(c-b)}[/MATH], we need to look for the common denominator, and in our case, the common denominator would be $(a-b)(b-c)(a-c)$, thus:

[MATH] \frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-c)(b-a)}+\frac{c^4}{(c-a)(c-b)}[/MATH]

[MATH]=\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-c)(-)(a-b)}+\frac{c^4}{(-)(a-c)(-)(b-c)}[/MATH]

[MATH]=\frac{a^4}{(a-b)(a-c)}-\frac{b^4}{(b-c)(a-b)}+\frac{c^4}{(a-c)(b-c)}[/MATH]

[MATH]=\frac{a^4(b-c)-b^4(a-c)+c^4(a-b)}{(a-b)(b-c)(a-c)}[/MATH]

It's like option B is the answer since

[MATH]\left.\begin{array}{l} a^4(b-c)-b^4(a-c)+c^4(a-b) \\ =a^4b-a^4c-ab^4+b^4c+c^4(a-b) \\ =a^4b-ab^4-a^4c+b^4c+c^4(a-b) \\ =ab(a^3-b^3)-c(a^4-b^4)+c^4(a-b) \\ =ab(a-b)(a^2+ab+b^2)-c(a^2-b^2)(a^2+b^2)+c^4(a-b) \\ =ab(a-b)(a^2+ab+b^2)-c(a-b)(a+b)(a^2+b^2)+c^4(a-b) \\ =(a-b)(ab(a^2+ab+b^2)-c(a+b)(a^2+b^2)+c^4) \\ =(a-b)(a^3b+a^2b^2+ab^3-(ac+bc)(a^2+b^2)+c^4) \\ =(a-b)(a^3b+a^2b^2+ab^3-(a^3c+ab^2c+a^2bc+b^3c)+c^4) \\ =(a-b)(a^3b+a^2b^2+ab^3-a^3c-ab^2c-a^2bc-b^3c+c^4) \\ =(a-b)(a^3b-a^3c+a^2b^2-a^2bc+ab^3-ab^2c-b^3c+c^4) \\ =(a-b)(a^3(b-c)+a^2b(b-c)+ab^2(b-c)-c(b^3-c^3)) \\ =(a-b)(a^3(b-c)+a^2b(b-c)+ab^2(b-c)-c(b-c)(b^2+bc+c^2)) \\ =(a-b)(b-c)(a^3+a^2b+ab^2-c(b^2+bc+c^2)) \\ =(a-b)(b-c)(a^3+a^2b+ab^2-b^2c-bc^2-c^3) \\ =(a-b)(b-c)(a^3-c^3+a^2b-bc^2+ab^2-b^2c) \\ =(a-b)(b-c)((a-c)(a^2+ac+c^2)+b(a^2-c^2)+b^2(a-c)) \\ =(a-b)(b-c)((a-c)(a^2+ac+c^2)+b(a+c)(a-c)+b^2(a-c)) \\ =(a-b)(b-c)(a-c)(a^2+ac+c^2+b(a+c))+b^2) \\ =(a-b)(b-c)(a-c)(a^2+b^2+c^2+ab+bc+ca)\end{array}\right\}[/MATH]19 steps

Now, let's study the option A, that is to replace [MATH]\color{yellow}\bbox[5px,purple]{c^4(a-b)}[/MATH] in

[MATH]\color{black}a^4(b-c)-b^4(a-c)+\color{yellow}\bbox[5px,purple]{c^4(a-b)}[/MATH] with [MATH]\color{yellow}\bbox[5px,purple]{c^4(a-c-b+c)}[/MATH]:

[MATH]\left.\begin{array}{l}\color{black}a^4(b-c)-b^4(a-c)+\color{yellow}\bbox[5px,purple]{c^4(a-c-b+c)} \\ =\color{black}a^4(b-c)-b^4(a-c)+\color{yellow}\bbox[5px,purple]{c^4(a-c)-c^4(b-c)} \\ =a^4(b-c)-c^4(b-c)-b^4(a-c)+c^4(a-c) \\ =(b-c)(a^4-c^4)-(a-c)(b^4-c^4) \\ =(b-c)(a^2-c^2)(a^2+c^2)-(a-c)(b^2-c^2)(b^2+c^2) \\ =(b-c)(a-c)(a+c)(a^2+c^2)-(a-c)(b-c)(b+c)(b^2+c^2) \\ =(b-c)(a-c)((a+c)(a^2+c^2)-(b+c)(b^2+c^2)) \\ =(b-c)(a-c)(a^3+ac^2+a^2c+c^3-b^3-bc^2-b^2c-c^3) \\ =(b-c)(a-c)(a^3-b^3+ac^2-bc^2+a^2c-b^2c) \\ =(b-c)(a-c)((a-b)(a^2+ab+b^2)+c^2(a-b)+c(a^2-b^2)) \\ =(b-c)(a-c)((a-b)(a^2+ab+b^2)+c^2(a-b)+c(a-b)(a+b)) \\ =(a-b)(b-c)(a-c)(a^2+ab+b^2+c^2+c(a+b)) \\ =(a-b)(b-c)(a-c)(a^2+b^2+c^2+ab+bc+ca)\end{array}\right\}[/MATH]12 steps

Of course we would select option A as it gives us easier and less work than the other form as stated in option B.

[MATH] a^2+b^2+c^2+ab+bc+ca[/MATH]

Question 6: What would you do if you have to add up the expression above, given $a=-\sqrt{6}+\sqrt{7}+\sqrt{8}),\,b=\sqrt{6}-\sqrt{7}+\sqrt{8},\,c=\sqrt{6}+\sqrt{7}-\sqrt{8}$?

A. By squaring $a,\,b$ and $c$ and also multiplying $ab$, $bc$ and $ca$ together before adding.

B. By using the formula $(a+b)^2=a^2+b^2+2ab$ to simplify it before adding.

Answer:

Both options are viable routes to evaluate [MATH] a^2+b^2+c^2+ab+bc+ca[/MATH], but it won't be easy to go with squaring $a,\,b$ and $c$ and multiplying $ab$, $bc$ and $ca$ together before adding because:

$a^2=(-\sqrt{6}+\sqrt{7}+\sqrt{8})^2$

$\,\,\,\,\,\,\,=(-\sqrt{6}+\sqrt{7})^2+2(-\sqrt{6}+\sqrt{7})(\sqrt{8})+(\sqrt{8})^2$

$\,\,\,\,\,\,\,=6-2\sqrt{6}\sqrt{7}+7-2\sqrt{6}\sqrt{8}+2\sqrt{7}\sqrt{8}+8$

$\,\,\,\,\,\,\,=21-2\sqrt{6}\sqrt{7}-2\sqrt{6}\sqrt{8}+2\sqrt{7}\sqrt{8}$

$b^2=(\sqrt{6}-\sqrt{7}+\sqrt{8})^2$

$\,\,\,\,\,\,\,=(\sqrt{6}-\sqrt{7})^2+2(\sqrt{6}-\sqrt{7})(\sqrt{8})+(\sqrt{8})^2$

$\,\,\,\,\,\,\,=6-2\sqrt{6}\sqrt{7}+7+2\sqrt{6}\sqrt{8}-2\sqrt{7}\sqrt{8}+8$

$\,\,\,\,\,\,\,=21-2\sqrt{6}\sqrt{7}+2\sqrt{6}\sqrt{8}-2\sqrt{7}\sqrt{8}$

$c^2=(\sqrt{6}+\sqrt{7}-\sqrt{8})^2$

$\,\,\,\,\,\,\,=(\sqrt{6}+\sqrt{7})^2+2(\sqrt{6}+\sqrt{7})(-\sqrt{8})+(-\sqrt{8})^2$

$\,\,\,\,\,\,\,=6+2\sqrt{6}\sqrt{7}+7-2\sqrt{6}\sqrt{8}-2\sqrt{7}\sqrt{8}+8$

$\,\,\,\,\,\,\,=21+2\sqrt{6}\sqrt{7}-2\sqrt{6}\sqrt{8}-2\sqrt{7}\sqrt{8}$

$ab+bc+ca$

$=a(b+c)+bc$

$=(-\sqrt{6}+\sqrt{7}+\sqrt{8})(\sqrt{6}-\sqrt{7}+\sqrt{8}+\sqrt{6}+\sqrt{7}-\sqrt{8})+(\sqrt{6}-\sqrt{7}+\sqrt{8})(\sqrt{6}+\sqrt{7}-\sqrt{8})$

$=(-\sqrt{6}+\sqrt{7}+\sqrt{8})(2\sqrt{6})+(\sqrt{6}-(\sqrt{7}-\sqrt{8}))(\sqrt{6}+(\sqrt{7}-\sqrt{8}))$

$=(-\sqrt{6}+\sqrt{7}+\sqrt{8})(2\sqrt{6})+((\sqrt{6})^2-(\sqrt{7}-\sqrt{8})^2)$

$=(-12+2\sqrt{6}\sqrt{7}+2\sqrt{6}\sqrt{8})+(6-(7-2\sqrt{7}\sqrt{8}+8))$

$=-12+2\sqrt{6}\sqrt{7}+2\sqrt{6}\sqrt{8}+(-9+2\sqrt{7}\sqrt{8})$

$=-21+2\sqrt{6}\sqrt{7}+2\sqrt{6}\sqrt{8}+2\sqrt{7}\sqrt{8}$

Therefore

[MATH] a^2+b^2+c^2+ab+bc+ca[/MATH]

[MATH]=21-2\sqrt{6}\sqrt{7}-2\sqrt{6}\sqrt{8}+2\sqrt{7}\sqrt{8}+21-2\sqrt{6}\sqrt{7}+2\sqrt{6}\sqrt{8}-2\sqrt{7}\sqrt{8}[/MATH]

[MATH]\,\,\,+21+2\sqrt{6}\sqrt{7}-2\sqrt{6}\sqrt{8}-2\sqrt{7}\sqrt{8}+(-21+2\sqrt{6}\sqrt{7}+2\sqrt{6}\sqrt{8}+2\sqrt{7}\sqrt{8})[/MATH]

$=42$

Compare it with using the identities $(a+b)^2=a^2+b^2+2ab,\,(b+c)^2=b^2+c^2+2bc,\,(a+c)^2=a^2+c^2+2ac$, notice that adding them up yields:

$(a+b)^2+(b+c)^2+(a+c)^2$

$\,\,\,=a^2+b^2+2ab+b^2+c^2+2bc+a^2+c^2+2ac$

$\,\,\,=2(a^2+b^2+c^2)+2(ab+bc+ac)$

Thus [MATH]\frac{(a+b)^2+(b+c)^2+(a+c)^2}{2}=a^2+b^2+c^2+ab+bc+ac[/MATH].

It means

[MATH]a^2+b^2+c^2+ab+bc+ac[/MATH]

[MATH]=\frac{(a+b)^2+(b+c)^2+(a+c)^2}{2}[/MATH]

[MATH]=\frac{(-\sqrt{6}+\sqrt{7}+\sqrt{8}+\sqrt{6}-\sqrt{7}+\sqrt{8})^2+(\sqrt{6}-\sqrt{7}+\sqrt{8}+\sqrt{6}+\sqrt{7}-\sqrt{8})^2}{2}[/MATH]

[MATH]\,\,\,+\dfrac{(-\sqrt{6}+\sqrt{7}+\sqrt{8}+\sqrt{6}+\sqrt{7}-\sqrt{8})^2}{2}[/MATH]

[MATH]=\frac{(2\sqrt{8})^2+(2\sqrt{6})^2+(2\sqrt{7})^2}{2}[/MATH]

[MATH]=\frac{32+24+28}{2}[/MATH]

[MATH]=\frac{84}{2}[/MATH]

$=42$

Question 7: What is the sum of the three fractions?

A. 22

B. 32

C. 42

Answer:

From our previous working, we know that 42 is the answer.

## No comments:

## Post a Comment