Question

6. If the perimeter of a triangle is given by 3x – y +6 and two of it sides are given by 4x-9 and 12y - X+8, then find its third side.
Answer step by step ​

Answer 1

Answer:

x = (√2+1)^1/3 - (√2–1)^1/3. ……………….(1).

Cubing both sides.

[ Formula : (a - b)^3 = a^3 - b^3 -3.a.b.(a+b).}

or, x^3 = (√2+1) -(√2–1) - 3.(√2+1)^1/3.(√2–1)^1/3.{(√2+1)^1/3-(√2–1)^1/3}

Putting {(√2+1)^1/3-(√2–1)^1/3} = x.

or, x^3 = √2+1-√2+1 -3.{ (√2)^2-(1)^1}^1/3. {x} .

or, x^3 = 2 - 3.(1)^1/3.x

or, x^3 = 2 - 3.x

or, x^3 +3.x = 2. Answer.

Answer 2

Answer:

x = (√2+1)^1/3 - (√2–1)^1/3. ……………….(1).

Cubing both sides.

[ Formula : (a - b)^3 = a^3 - b^3 -3.a.b.(a+b).}

or, x^3 = (√2+1) -(√2–1) - 3.(√2+1)^1/3.(√2–1)^1/3.{(√2+1)^1/3-(√2–1)^1/3}

Putting {(√2+1)^1/3-(√2–1)^1/3} = x.

or, x^3 = √2+1-√2+1 -3.{ (√2)^2-(1)^1}^1/3. {x} .

or, x^3 = 2 - 3.(1)^1/3.x

or, x^3 = 2 - 3.x

or, x^3 +3.x = 2. Answer.

Answer 3

Answer:

x = (√2+1)^1/3 - (√2–1)^1/3. ……………….(1).

Cubing both sides.

[ Formula : (a - b)^3 = a^3 - b^3 -3.a.b.(a+b).}

or, x^3 = (√2+1) -(√2–1) - 3.(√2+1)^1/3.(√2–1)^1/3.{(√2+1)^1/3-(√2–1)^1/3}

Putting {(√2+1)^1/3-(√2–1)^1/3} = x.

or, x^3 = √2+1-√2+1 -3.{ (√2)^2-(1)^1}^1/3. {x} .

or, x^3 = 2 - 3.(1)^1/3.x

or, x^3 = 2 - 3.x

or, x^3 +3.x = 2. Answer.

Answer 4

Answer:

x = (√2+1)^1/3 - (√2–1)^1/3. ……………….(1).

Cubing both sides.

[ Formula : (a - b)^3 = a^3 - b^3 -3.a.b.(a+b).}

or, x^3 = (√2+1) -(√2–1) - 3.(√2+1)^1/3.(√2–1)^1/3.{(√2+1)^1/3-(√2–1)^1/3}

Putting {(√2+1)^1/3-(√2–1)^1/3} = x.

or, x^3 = √2+1-√2+1 -3.{ (√2)^2-(1)^1}^1/3. {x} .

or, x^3 = 2 - 3.(1)^1/3.x

or, x^3 = 2 - 3.x

or, x^3 +3.x = 2. Answer.