(a+x)2/3+4 (a-x) 2/3 - 5 (a²-x2) 1/3
Answers
Solution :-
→ (a + x)⅔ + 4(a - x)⅔ = 5(a² - x²)⅓
→ (a + x)⅔ + 4(a - x)⅔ - 5(a² - x²)⅓ = 0
→ (a + x)⅔ + 4(a - x)⅔ - 5(a + x)⅓(a - x)⅓ = 0
→ [(a + x)⅓]² + [2(a - x)⅓]² - 2*2(a + x)⅓(a - x)⅓ - (a + x)⅓(a - x)⅓ = 0
using a² + b² - 2ab = (a - b)²
→ [(a + x)⅓ - 2(a - x)⅓]² - (a + x)⅓(a - x)⅓ = 0
→ [a⅓ - 2a⅓ + x⅓ - 2x⅓]² - (a + x)⅓(a - x)⅓ = 0
→ (- a⅓ - x⅓)² - (a + x)⅓(a - x)⅓ = 0
→ [(-1)(a⅓ + x⅓)]² - (a + x)⅓(a - x)⅓ = 0
→ (a⅓ + x⅓)² - [(a + x)(a - x)]⅓ = 0
→ (a⅓ + x⅓)² - (a² - x²)⅓ = 0
Lets try to solve with another method,
→ (a + x)⅔ + 4(a - x)⅔ = 5(a² - x²)⅓
cube both sides,
→ [(a + x)⅔ + 4(a - x)⅔]³ = [5(a² - x²)⅓]³
using (a + b)³ = a³ + b³ + 3ab(a + b) in LHS,
→ {(a + x)⅔}³ + {4(a - x)⅔}³ + 3(a + x)⅔ * 4 * (a - x)⅔[(a + x)⅔ + 4(a - x)⅔] = 125(a² - x²)
→ (a + x)² + 64(a - x)² + 12 * (a + x)⅔ * (a - x)⅔ * 5 * (a² - x²)⅓ = 125(a² - x²)
→ (a + x)² + 64(a - x)² + 60{(a + x)(a - x)}⅔ * (a² - x²)⅓ = 125(a² - x²)
→ (a + x)² + 64(a - x)² + 60(a² - x²)⅔ * (a² - x²)⅓ = 125(a² - x²)
→ (a + x)² + 64(a - x)² + 60(a² - x²)^(2/3 + (1/3) = 125(a² - x²)
→ (a + x)² + 64(a - x)² + 60(a² - x²) = 125(a² - x²)
→ (a + x)² + 64(a - x)² = 125(a² - x²) - 60(a² - x²)
→ (a + x)² + 64(a - x)² = 65(a² - x²)
→ (a + x)² + 64(a - x)² = 64(a² - x²) + (a² - x²)
→ (a + x)² - (a² - x²) = 64(a² - x²) - 64(a - x)²
→ (a + x)[(a + x) - (a - x)] = 64(a - x)[(a + x) - (a + x)]
→ (a + x) * 0 = 64(a - x) * 0
→ 0 = 0
Learn more :-
JEE mains Question :-
https://brainly.in/question/22246812
. Find all the zeroes of the polynomial x4
– 5x3 + 2x2+10x-8, if two of its zeroes are 4 and 1.
https://brainly.in/question/39026698
Answer:
0
Step-by-step explanation:
→ (a + x)3 + 4(a - x)² = 5(a² - x²)3
→ (a + x)3 + 4(a - x)² - 5(a² - x²)3 = 0
(a + x)+4(a - x) - 5(a + x)%(a - x) =
→ [(a + x)]² + [2(a - x)]² - 2*2(a + x)(a - x) - (a + x)(a - x) = 0 using a² + b² - 2ab = (a - b)²
→ [(a + x) - 2(a - x)]² - (a + x)(a - x) =
→ [aз - 2a + x3 - 2׳]² - (a + x)(a - X)
= 0
→ (-a½ - X³)² - (a + x)(a - x) = 0 ->
→ [(-1)(a½ + ׳)]² - (a + x)(a - x) = 0
→ (a + x)² -[(a + x)(a - x)] = 0
→ (a + x)² - (a² - x²) = 0