Question

If given in the following expression, find the value of

(x^3 + y^3):(x^3-y^3)
If (x^2+y^2)/(x^2 -y^2)= 17 /8​

Answer 1

Explanation:

Given that:

{(x²+y²)/(x²-y²)} = 17/8

On applying cross multiplication then

since, (a/b) = (c/d) ⇛a(d) = b(c)

Where, x = x²+y², b = x²-y², c = 17 and d = 8

So,

⇛17(x²-y²) = 8(x²+y²)

Multiply the number outside of the brackets with variables in the brackets on both LHS and RHS.

⇛17x²-17y² = 8x²+8y²

⇛17x²-17y² - 8x²-8y² = 0

⇛17x²- 8x² - 17y² - 8y² = 0

⇛9x² - 25y² = 0

⇛9x² = 25y²

⇛9/25 = y²/x²

⇛25/9 = x²/y²

⇛x/y = √(25/9)

⇛x/y = √{(5*5)/(3*3)}

⇛x/y = 5/3

Therefore, x:y = 5:3

Let x = 5a and y = 3a

Now, (x³+y³):(x³-y³)

= {(x³+y³)/(x³-y³)} = [{(x+y)(x²-xy+y²)}/{(x-y)(x²+x-y+y²)}]

= [{(5a)³ + (3a)³}/{(5a)³ - (3a)³}]

= [{(5*5*5*a*a*a) + (3*3*3*a*a*a)}/{(5*5*5*a*a*a) - (3*3*3*a*a*a)}]

= {(125a³ + 27a³)/(125a³ - 27a²)}

= {(152a³)/(98a³)}

= 152/98

Write the obtained fraction in lowest form by cancelling method.

= {(152÷2)/(98÷2)}

= 76/49

Therefore, , (x³+y³):(x³-y³) = 76/49

Answer: Hence, the required value of (x³+y³):(x³-y³) is 76/49.

Please let me know if you have any other questions.

Answer 2

76:49 or 49:76

Explanation:

The second relation gives an expression for y in terms of x:

`(x^2+y^2)/(x^2-y^2)=(17)/(8)\\\\8x^2+8y^2=17x^2-17y^2\qquad\text{cross multiply}\\\\25y^2=9x^2\qquad\text{add $(17y^2-8x^2)$}\\\\5y=\pm3x\qquad\text{take the square root}\\\\y=\pm(3)/(5)x\qquad\text{divide by 5}`

Substituting into the expression we want the value of, we can factor out x^3 to get ...

`(x^3+y^3):(x^3-y^3)=(1\pm(3/5)^3):(1\mp(3/5)^3)=(125\pm27):(125\mp27)\\\\=152:98\text{ or }98:152=\boxed{76:49\text{ or }49:76}\m`

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Additional comment

There is nothing in the given requirement that means x and y must have the same sign. If the signs are different, then the value of the expression is less than 1, rather than greater than 1.