Question

Factorise 125m^3-343n^3​

Answer 1

$\huge\color{indigo}\boxed{\colorbox{lime}{solution: ♞ }}$

$(5m - 7n). (25m² + 35mn + 49n²)$

$\color{fuchsia}Steps:$

$STEP1: \: Equation \: at \: the \: end \: of \: step 1</p><p></p><p>$

$= > \: (125 .(m3)) - {7}^{3} {n}^{3}$

$STEP 2 :</p><p>Equation \: at \: the \: end \: of \: step2:</p><p></p><p>$

$= > {5}^{3} {m}^{3} - {7}^{3} {n}^{3}$

$STEP3:</p><p></p><p>Trying \: to \: factor \: as \: \: a \\ Difference \: of \: Cubes:</p><p></p><p>$

$.Theory : A \: difference \: of \: two \: perfect \\ cubes, \: {a}^{3}  -  {b}^{3} \: can \: be \: factored \: into : \\ </p><p>              (a-b) • ( {a}^{2}  +ab + {b}^{2} ) \\ Proof : \: (a-b)•( {a}^{2} +ab+ {b}^{2} )  = \\ {a}^{3} + {a}^{2} b+a {b}^{2} -b {a}^{2} - {b}^{2} a- {b}^{3}  = \\ {a}^{3} +( {a}^{2} b-b {a}^{2} )+(a {b}^{2} - {b}^{2} a)- {b}^{3} = \\ {a}^{3} +0+0- {a}^{3}  = \\ {a}^{3} - {b}^{3} \\ Check : \: 125 \: is \: the \: cube \: of \: 5. \\ Check : \: 343 \: is \: the \: cube \: of \: 7. \\ Check :  {m}^{3} \: is \: the \: cube \: of {m}^{1}. \\ Check :  {n}^{3} \: is \: the \: cube \: of \: {n}^{1}. \\ </p><p>Factorization \: is : \\ \: \: \: \: \: (5m - 7n)  •  (25 {m}^{2}  + 35mn + 49 {n}^{2} )$

$\color{blue}Try\: to \: factor \: a \: multi \: variable \\ \color{blue} polynomial : </p><p></p><p>$

$\color{red}⚠factorisation \: fails⚠$

$\color{33cc99}So, \: the \: final \: answer \: is, \\\color{33cc99} (5m - 7n).(25 {m}^{2} + 35mn + 49 {n}^{2} )$