Question

What's the lowest common multiple of (a³ + b³) and (a⁴ - b⁴).

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Answer 1

$\;\large\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

The question is about finding the lowest common multiple (LCM) of (a³ + b³) and (a⁴ - b⁴). It is easy to find the LCM if the both given brackets. To find the lowest common multiple of those numbers, we need to use some properties and basic maths calculation.

Solution:

Using the property,

• a³ + b³ = (a + b)(a² - ab + b²).

So,

⟹ a³ + b³ = (a + b)(a² - ab + b²).

⟹ a⁴ - b⁴ = (a² - b²)(a² + b²).

⟹ (a - b)(a + b)(a² + b²).

Now,

LCM = ?

⟹ (a + b)(a - b).

⟹ (a² + b²)(a² - ab + b²).

⟹ (a⁴ + b⁴)(a² - ab + b²).

⟹ (a³ + b³)(a - b)(a² + b²).

Answer 2

$\Large{ \underbrace{ \underline{ \sf{Understanding \: the \: Question}}}}$

Here, this is a question from Algebric expressions where we have to find Lowest common multiple (LCM) of the given complex terms.

So obviously we can't find LCM of complex terms, we have to first expand the given terms into their factors by using Identities.

So let's start!

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a⁴-b⁴

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=> a⁴-b⁴= (a²)²-(b²)²

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=> a⁴-b⁴= (a²+b²)(a²-b²)

[★ a²-b²=(a+b)(a-b)]

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=> a⁴-b⁴=(a²+b²)(a+b)(a-b)..(1)

[★ a²-b²=(a+b)(a-b)]

$\rule{300}{2}$

a³+b³

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=> a³+b³=(a+b)(a²+b²-ab)...(2)

[•°• Identity used]

$\rule{300}{2}$

Here equation (1) and (2) are the expanded forms of given terms.

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So let's find the LCM of expanded terms.

[See the attachment]

$\rule{300}{2}$

So we can conclude that LCM of (a⁴-b⁴) and (a³+b³) is:

(a+b)×(a²+b²)×(a-b)×(a²+b²-ab)

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Now we have to simplify this term to obtain the final answer.

(a+b)(a²+b²)(a-b)(a²+b²-ab)

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=> (a²+b²)(a-b)(a+b)(a²+b²-ab)

=> (a²+b²)(a-b)(a³+b³)

[★(a+b)(a²+b²-ab)=a³+b³]

Hence LCM of (a³+b³) and (a⁴-b⁴) is:

• (a²+b²)(a-b)(a³+b³)

$\rule{300}{2}$

MORE RELATED FORMULAE:

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$

$\rule{300}{2}$

NOTE- Kindly see this answer on web for better understanding.