Question

☺ Solve it ☺

☺ Explanation Needed ​

Answer 1

Given :

• x³ + 64y³
• 125p³ + q³
• 125k³ + 27m³
• 2l³ + 432m³

Solution :

Apply identity : a³ + b³ = (a + b)(a² - ab + b²)

• x³ + 64y³

→ (x)³ + (4y)³

→ (x + 4y)[(x)² - x*4y + (4y)²]

→ (x + 4y)[x² - 4xy + 16y²]

• 125p³ + q³

Apply identity : a³ + b³ = (a + b)(a² - ab + b²)

→ (5p)³ + (q)³

→ (5p + q)[(5p)² - 5p*q + (q)²]

→ (5p + q)[25p² - 5pq + q²]

• 125k³ + 27m³

Apply identity : a³ + b³ = (a + b)(a² - ab + b²)

→ (5k)³ + (3m)³

→ (5k + 3m)[(5k)² - 5k*3m + (3m)²]

→ (5k + 3m)[25k² - 15km + 9m²]

• 2l³ + 432m³

Apply identity : a³ + b³ = (a + b)(a² - ab + b²)

Take 2 as a common

→ 2[l³ + 216m³]

→ 2[(l)³ + (6m)³]

→ 2[{l + 6m}{(l)² - l*6m + (6m)²}]

→ 2[{l + 6m}{l² - 6lm + 36m²}]

Answer 2

$\bigstar$ Given

$\rm 1 .\ x^3+64y^3\\\\\rm 2.\ 125p^3+q^3\\\\\rm 3.\ 125k^3+27m^3\\\\\rm 4.\ 2l^3+432^3$

$\bigstar$ To Find

Factorization of given equations

$\bigstar$ Needs to know

$\bf a^3+b^3=(a+b)(a^2-ab+b^2)\ \bigstar$

$\bigstar$ Solution

$\rm 1.\ x^3+64y^3\\\\\rightarrow \rm x^3+(4y)^3\\\\\rightarrow \rm (x+4y)(x^2-x(4y)+(4y)^2)\\\\\rightarrow \rm (x+4y)(x^2-4xy+16y^2)$

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$\rm 2.\ 125p^3+q^3\\\\\rm \rightarrow (5p)^3+q^3\\\\\rm \rightarrow (5p+q)((5p)^2-(5p)q+q^2)\\\\\rm \rightarrow (5p+q)(25p^2-5pq+q^2)$

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$3.\ \rm 125k^3+27m^3\\\\\rm \rightarrow (5k)^3+(3m)^3\\\\\rm \rightarrow (5k+3m)((5k)^2 -(5k)(3m)+(3m)^2)\\\\\rm \rightarrow (5k+3m)(25k^2-15km+9m^2)$

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$4.\ \rm 2l^3+432m^3\\\\\rm \rightarrow 2[l^3+216m^3]\\\\\rm \rightarrow 2[l^3+(6m)^3]\\\\\rm \rightarrow 2[(l+6m)(l^2-l(6m)+(6m)^2)]\\\\\rm \rightarrow 2[(l+6m)(l^2-6lm+36m^2)]$

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