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1)If (x+y)^2 = 31 and xy=6, what is the value of (x-y)^2?]

2)Compute the product $\frac{(1998^2 - 1996^2)(1998^2 - 1995^2) \cdots (1998^2 - 0^2)}{(1997^2 - 1996^2)(1997^2 - 1995^2) \cdots (1997^2 - 0^2)}.$

3)A square park measures 170 feet along each side. Two paved paths run from each corner to the opposite corner and extend 3 feet inwards from each corner, as shown. What is the total area, in square feet, taken by the paths?

4)If $\left(a-\frac 1a\right)^2=4$ and $a > 1,$ what is the absolute value of $a^3 - \frac1{a^3}?$

5)If $a+b=7$ and $a^3+b^3=42$, what is the value of the sum $\dfrac{1}{a}+\dfrac{1}{b}$? Express your answer as a common fraction.

Sep 14, 2019

#1
+2

1.)

$$\text{If you want to figure out }(x-y)^2\text{, You can expand it to }x^2-2xy+y^2$$

$$\text{Notice how }xy=6\text{ now that means we can place it into our } x^2-2xy+y^2\text{ equation}$$

$$\text{So we have: }x^2-2xy+y^2\Rightarrow{x^2}-2(6)+y^2\Rightarrow{\color{green}x^2+y^2-12}$$

$$\text{Now consider one of the equations we were given: }(x+y)^2=31$$

$$\text{We can now expand the following: } (x+y)^2=31\Rightarrow{\color{red}x^2+2xy+y^2=31}$$

$$\text{Now we place }xy=6\text{ into the red equation}$$

$$x^2+2xy+y^2=31\rightarrow{x^2}+2(6)+y^2=31\rightarrow{\color{red}x^2+y^2=19}$$

$$\text{Now we combine the new red equation and the green equation together for the answer}$$

$${\color{red}x^2+y^2=19}\text{ and }{\color{green}x^2+y^2-12}\Rightarrow{19-12}=\boxed{7}$$

I do not have time to answer the other questions, but I am sure that the all involve expanding special cases and combining equations!

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Sep 14, 2019
edited by CalculatorUser  Sep 14, 2019
edited by CalculatorUser  Sep 14, 2019
#2
+2

$$4)If \ \left(a-\frac 1a\right)^2=4\ and \ a > 1, what is the absolute value of a^3 - \frac1{a^3}?$$

Expand (a -1/a)^2  =  4   →   a^2 - 2  + 1/a^2   = 4  →  a^2 + 1/a^2  =  6   (1)

And  (a - 1/a)^2  = 4       take the positive root

a - 1/a  = 2      (2)

Factor     a^3  - 1/a^3    as  a difference of cubes

(a  -1/a)  ( a^2 + 1  +  1/a^2)

(a - 1/a) ( [a^2 + 1/a^2]  + 1)      (3)

Sub (1) and (2)  into (3)

(2)  ( 6 + 1)

(2) (7)  =

14   Sep 14, 2019
#3
+2

$$5)If a+b=7 and a^3+b^3=42, what is the value of the sum \dfrac{1}{a}+\dfrac{1}{b}?$$

a  +  b  =    7     square both sides

a^2 + 2ab +b^2  = 49         (1)

a^3 + b^3  =  42         factor the left side as a sum of cubes

(a + b) (a^2  - ab + b^2)  = 42

(7) (a^2 - ab + b^2)  = 42          divide both sides by 7

a^2 - ab + b^2  =  6       (2)

Subtract  (2)  from (1)   and we have that

3ab  =  43

ab = 43/3        ( 3)

And  we can simplify    1/a  +  1/b   as     (a + b)  /  ab

So

1/a   +  1/b   =

(a + b)  / ab    =

7  /  ( 43/3)   =

(7/1)  / (43/3)  =

(7/1) *  (3/43)  =

7 * 3  / 43   =

21 / 43   Sep 14, 2019
edited by CPhill  Sep 14, 2019
#4
+2

(1998^2 - 1996^2)(1998^2 - 1995^2)   ........1998^2

_________________________________________________________

(1997^2  - 1996^2) ( 1997^2 - 1995^2)  .......1997^2

(1998 + 1996)(1998 - 1996) ( 1998 + 1995)(1998 - 1995) (1998 + 1994) (1998 - 1994).....1998^2

_____________________________________________________________________________

(1997 + 1996) (1997 - 1996) (1997 + 1995)(1997 - 1995) (1997 + 1994)(1997 - 1994).....1997^2

[ (3994)(2)]  [ (3993)(3) ] [(3992)(4) ].......[(1998)(1998)]

____________________________________________

[ (3993)(1)] [ (3992)(2)]  [ (3991)(3)]..........[(1997)(1997)]

Note that  each pair of products  in the numerator sum to 3996  and each pair of of products in the numerator sum to 3994

After cancellation of like terms we are left with

3994  *     ( 4)                 [(1998)(1998)]

_____   .....     _____________

( 3991)            [ (1997)(1997)]

And....note that the sum of the non-cancelled terms in the middle fraction's numerator and denominator  =   3995

This implies.....that after complete evaluation ( and cancellations), we will be left with

3994   *    (1998)

______  =

(1997)

(2)(1997) * (1998)

_______________   =

(1997)

2 * 1998   =

3996   Sep 14, 2019
edited by CPhill  Sep 14, 2019