Question

${2l}^{3 \:} + {432m}^{3}$
please help for this question​

Answer 1

$\huge\tt\colorbox{pink}{Answer}$

$2 {l}^{3} + 432 {m}^{3} \\ = 2( {l}^{3} + 216 {m}^{3} ) \\ = 2( {l}^{3} + {(6m)}^{3} ) \\ = 2(l + 6m)( {l}^{2} +( 6 {m)}^{2} - l \times 6m) \\ = 2(l + 6m)( {l}^{2} + 36 {m}^{2} - 6lm)$

Answer 2

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$\large \sf \red{2 {l}^{3} + 432 {m}^{3} } = (2l + 432m {)}^{3}$

To solve this, we will use the identity:-

(a+b)³=a³+3x²y+3xy²+y³

Here a= 2l, b=432m

= 2l³+ 3(2l)²(432m) +3(2l)(432m)²+432m³

=8l³+3(8l)²(432m)+ 6l(186624m)²+ 80621568m³

=8l³+24l²(432m)+1119744m²+80621568m³

=8l³+10368l²m+1119744m²+80621568m³

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