64 - p^6
You should definitely memorize the factoring of a difference of squares, and it would also be good to know the sum/difference of cubes.
a^2 - b^2 = (a+b)(a-b)
a^3 + b^3 = (a+b)(a^2 - ab+ b^2)
8^2 - (p^3)^2 = (8 + p^3)(8-p^3)
= (2+p)(4-2p+p^2)(2-p)(4+2p+p^2)
= - (p^6 - 64)
= - (p^3 + 8)(p^3 - 8)
= - (p + 2)(p^2 - 2p + 4)(p - 2)(p^2 + 2p + 4) Answer//
I use to love math not anymore
Before starting
(a + b)³ = (a + b)².(a + b)
(a + b)³ = (a² + 2ab + b²).(a + b)
(a + b)³ = a³ + a²b + 2a²b + 2ab² + ab² + b³
(a + b)³ = a³ + b³ + 3a²b + 3ab²
(a + b)³ = (a³ + b³) + (3a²b + 3ab²)
(a + b)³ = (a³ + b³) + 3ab.(a + b)
a³ + b³ = (a + b)³ - 3ab.(a + b)
a³ + b³ = (a + b).[(a + b)² - 3ab]
a³ + b³ = (a + b).[a² + 2ab + b² - 3ab]
a³ + b³ = (a + b).(a² - ab + b²) → by replacing a by 2 and b by p
2³ + p³ = (2 + p).(4 - 2p + p²)
2³ + p³ = (2 + p).(p² - 2p + 4) ← memorize this result as (1)
(a - b)³ = (a - b)².(a - b)
(a - b)³ = (a² - 2ab + b²).(a - b)
(a - b)³ = a³ - a²b - 2a²b + 2ab² + ab² - b³
(a - b)³ = a³ - b³ - 3a²b + 3ab²
(a - b)³ = (a³ - b³) - (3a²b - 3ab²)
(a - b)³ = (a³ - b³) - 3ab.(a - b)
a³ - b³ = (a - b)³ + 3ab.(a - b)
a³ - b³ = (a - b)².(a - b) + 3ab.(a - b)
a³ - b³ = (a - b).[(a - b)² + 3ab]
a³ - b³ = (a - b).[a² - 2ab + b² + 3ab]
a³ - b³ = (a - b).(a² + ab + b²) → by replacing a by 2 and b by p
2³ - p³ = (2 - p).(4 + 2p + p²)
2³ - p³ = (2 - p).(p² + 2p + 4) ← memorize this result as (2)
= 64 - p⁶
= 2⁶ - p⁶
= (2³)² - (p³)² → recall: a² - b² = (a + b).(a - b)
= (2³ + p³).(2³ - p³) → recall (1)
= [(2 + p).(p² - 2p + 4)].(2³ - p³) → recall (2)
= [(2 + p).(p² - 2p + 4)].[(2 - p).(p² + 2p + 4)]
= (2 + p).(p² - 2p + 4).(2 - p).(p² + 2p + 4)
= (2 + p).(2 - p).(p² - 2p + 4).(p² + 2p + 4)
Thanks all. .............................
64-p^6
=
(2^3)^2-(p^3)^2
(2+p)(2-p)(4-2p+p^2)(4+2p-p^2)
64 = 2^6
Hence
2^6 - p^6 =>
(2^3)^2 - (p^3)^2 = >
(2^3 - p^3)(2^3 + p^3) =>
(2- p)(2^2 + 2p + p^2)(2 + p)(2^2 - 2p + p^2)
64 - p^6 = 2^6 - p^6 = -(p - 2) (p + 2) (p^2 - 2 p + 4) (p^2 + 2 p + 4)
Factor it as a difference of squares first:
(2³)² - (p³)² = (2³ - p³)(2³ + p³)
Then factor the sum and difference of cubes next.
plus or minus 2
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Answers & Comments
You should definitely memorize the factoring of a difference of squares, and it would also be good to know the sum/difference of cubes.
a^2 - b^2 = (a+b)(a-b)
a^3 + b^3 = (a+b)(a^2 - ab+ b^2)
8^2 - (p^3)^2 = (8 + p^3)(8-p^3)
= (2+p)(4-2p+p^2)(2-p)(4+2p+p^2)
64 - p^6
= - (p^6 - 64)
= - (p^3 + 8)(p^3 - 8)
= - (p + 2)(p^2 - 2p + 4)(p - 2)(p^2 + 2p + 4) Answer//
I use to love math not anymore
Before starting
(a + b)³ = (a + b)².(a + b)
(a + b)³ = (a² + 2ab + b²).(a + b)
(a + b)³ = a³ + a²b + 2a²b + 2ab² + ab² + b³
(a + b)³ = a³ + b³ + 3a²b + 3ab²
(a + b)³ = (a³ + b³) + (3a²b + 3ab²)
(a + b)³ = (a³ + b³) + 3ab.(a + b)
a³ + b³ = (a + b)³ - 3ab.(a + b)
a³ + b³ = (a + b).[(a + b)² - 3ab]
a³ + b³ = (a + b).[a² + 2ab + b² - 3ab]
a³ + b³ = (a + b).(a² - ab + b²) → by replacing a by 2 and b by p
2³ + p³ = (2 + p).(4 - 2p + p²)
2³ + p³ = (2 + p).(p² - 2p + 4) ← memorize this result as (1)
(a - b)³ = (a - b)².(a - b)
(a - b)³ = (a² - 2ab + b²).(a - b)
(a - b)³ = a³ - a²b - 2a²b + 2ab² + ab² - b³
(a - b)³ = a³ - b³ - 3a²b + 3ab²
(a - b)³ = (a³ - b³) - (3a²b - 3ab²)
(a - b)³ = (a³ - b³) - 3ab.(a - b)
a³ - b³ = (a - b)³ + 3ab.(a - b)
a³ - b³ = (a - b)².(a - b) + 3ab.(a - b)
a³ - b³ = (a - b).[(a - b)² + 3ab]
a³ - b³ = (a - b).[a² - 2ab + b² + 3ab]
a³ - b³ = (a - b).(a² + ab + b²) → by replacing a by 2 and b by p
2³ - p³ = (2 - p).(4 + 2p + p²)
2³ - p³ = (2 - p).(p² + 2p + 4) ← memorize this result as (2)
= 64 - p⁶
= 2⁶ - p⁶
= (2³)² - (p³)² → recall: a² - b² = (a + b).(a - b)
= (2³ + p³).(2³ - p³) → recall (1)
= [(2 + p).(p² - 2p + 4)].(2³ - p³) → recall (2)
= [(2 + p).(p² - 2p + 4)].[(2 - p).(p² + 2p + 4)]
= (2 + p).(p² - 2p + 4).(2 - p).(p² + 2p + 4)
= (2 + p).(2 - p).(p² - 2p + 4).(p² + 2p + 4)
Thanks all. .............................
64-p^6
=
(2^3)^2-(p^3)^2
=
(2+p)(2-p)(4-2p+p^2)(4+2p-p^2)
64 = 2^6
Hence
2^6 - p^6 =>
(2^3)^2 - (p^3)^2 = >
(2^3 - p^3)(2^3 + p^3) =>
(2- p)(2^2 + 2p + p^2)(2 + p)(2^2 - 2p + p^2)
64 - p^6 = 2^6 - p^6 = -(p - 2) (p + 2) (p^2 - 2 p + 4) (p^2 + 2 p + 4)
Factor it as a difference of squares first:
(2³)² - (p³)² = (2³ - p³)(2³ + p³)
Then factor the sum and difference of cubes next.
plus or minus 2