*Answer:– Answer: x = 1, x =-8 Proof:- The given equation:- x⅔ + x⅓ – 2 = 0 can be written as (x⅓)² + x⅓ – 2 = 0 ………………………………………………….(1) If y = x⅓ then (x⅓)² = y² and (1) → y² + y – 2 = 0 Or, (y² – 1) + y – 1 = 0 Factorising the quantity within brackets, (y+1) (y-1) + (y-1) = 0 Or, (y-1) (y+1 + 1) =0 [the common factor y-1 is taken out] Or, (y – 1) (y + 2) = 0 Or, y – 1 = 0, y + 2 = 0 ⇒ y = 1 , y = -2 ⇒ x⅓ = 1, x⅓ = -2 (∵ y = x⅓) Cubing (x⅓)³ = 1³ , (x⅓)³ = (-2)³ ⇒ x = 1, x = -8 (Answer) Verification:- x = 1 L.H.S. of (1) = (x⅓)² + x⅓ – 2 = (1⅓)² + 1⅓ – 2 = 1²+ 1 – 2 = 1² + 1 – 2 = 1+1 = 0 = R.H.S. x = -8 L.H.S. of (1) = (-8⅓)² + (-8)⅓ – 2 = (-2)² – 2 – 2 = 4 – 2 – 2 = 4 – 4 = 0 = R.H.S. [tex]hope \: it \: helps \: you[/tex] Reply
*Answer:–
Proof:-
The given equation:-
x⅔ + x⅓ – 2 = 0
can be written as (x⅓)² + x⅓ – 2 = 0 ………………………………………………….(1)
If y = x⅓ then (x⅓)² = y² and (1) →
y² + y – 2 = 0
Or, (y² – 1) + y – 1 = 0
Factorising the quantity within brackets,
(y+1) (y-1) + (y-1) = 0
Or, (y-1) (y+1 + 1) =0 [the common factor y-1 is taken out]
Or, (y – 1) (y + 2) = 0
Or, y – 1 = 0, y + 2 = 0
⇒ y = 1 , y = -2
⇒ x⅓ = 1, x⅓ = -2 (∵ y = x⅓)
Cubing (x⅓)³ = 1³ , (x⅓)³ = (-2)³
⇒ x = 1, x = -8 (Answer)
Verification:-
x = 1
L.H.S. of (1) = (x⅓)² + x⅓ – 2 = (1⅓)² + 1⅓ – 2 = 1²+ 1 – 2 = 1² + 1 – 2 = 1+1 = 0 = R.H.S.
x = -8
L.H.S. of (1) = (-8⅓)² + (-8)⅓ – 2 = (-2)² – 2 – 2 = 4 – 2 – 2 = 4 – 4 = 0 = R.H.S.
[tex]hope \: it \: helps \: you[/tex]