Answer:
The value of a⁶ + 6a²b² + b⁶ -1 = 7.
Step-by-step explanation:
<u>Correct</u><u> </u><u>question</u><u>:</u> Find the value of a⁶ + 6a²b² + b⁶ -1 = ? If a²+b² = 2
<u>Explanation:</u>
Given that: a² + b² = 2 – – – Equation(1)
On cubing both sides then
⇛(a² + b²)³ = 2³
Compare the LHS with (x + y)³, we get
x = a² and y = b²
Using identity (x + y)³ = x³ + y³ + 3xy(x+y), we have
⇛(a² + b²)³ = 2³
⇛(a²)³ + (b²)³ + 3a²b² (a² + b²) = 2*2*2
⇛a⁶ + b⁶ + 3a²b² (a² + b²) = 8
Since, from equation (1) a²+b² = 2 (Given)
⇛a⁶ + b⁶ + 3a²b² (2) = 8
⇛a⁶ + b⁶ + 6a²b² = 8
On subtracting 1 from both sides
⇛a⁶ + b⁶ + 6a²b² - 1 = 8-1
Therefore, a⁶ + 6a²b² + b⁶ -1 = 7
<u>Answer</u><u>:</u> Hence, the value of a⁶ + 6a²b² + b⁶ -1 = 7.
Please let me know if you have any other questions.