Math, asked by evangelinedilbet1, 1 year ago

Factorise:(p–q)3+(q–r)3+(r–p)3

Answers

Answered by pranjalgora9
0

)Answer:

Step-by-step explanation:

it can be written as:

3(p-q+q-r+r-p)  {as x*y + x* z + x*b can written as x(y+z+b) put x=3, y= p-q,z=q-r,b=r-p}

=3(0          

=0

thankyou

Answered by rishu6845
0

Answer:

\boxed{\pink{\large{3 \: ( \: p - q \: ) \: ( \: q - r \: ) \: ( \: r - p \: )}}}

Step-by-step explanation:

\bold{\underline{\green{To \: find}}}\longrightarrow \\ factors \: of \:  \\  {(p - q)}^{3}  +  {(q - r)}^{3}  +  {(r - p)}^{3}

\bold{\underline{\blue{Concept \: used}}}\longrightarrow \\ if \: (a + b + c) = 0 \: then \\  {a}^{3}  +  {b}^{3}  +  {c}^{3}   = 3 \: a \: b \: c

\bold{\underline{\red{Solution}}}\longrightarrow \\  {(p - q)}^{3}  +  {(q - r)}^{3}  +  {(r - p)}^{3}  \\ let \\ p - q = a \\ q - r = b \\ r - p = c \\ now \\ a + b + c = p - q + q - r + r - p \\  =  > a + b + c = 0 \\ so \\  {a}^{3}  +  {b}^{3}  +  {c}^{3}   =  3abc \\ now \: putting \: value \: of \: a \: b \: and \: c \\  =  > \pink{ {(p - q)}^{3}  +  {(q - r)}^{3}  +  {(r - p}^{3}  = 3(p - q) \: (q - r) \: (r - p) }\\\bold{\underline{\red{ Additional \: information}}}\longrightarrow \\ 1) \blue{{a}^{2}  -  {b}^{2}  = (a + b) \: (a - b)} \\ 2)\green{ {(a + b)}^{2}  =  {a}^{2}  +  {b}^{2}  + 2ab} \\ 3) \blue{{(a - b)}^{2}  =  {a}^{2}  +  {b}^{2}  - 2ab} \\ 4) \green{{a}^{3}  +  {b}^{3}  = (a + b) \: ( {a}^{2}  +  {b}^{2}  - ab)} \\ 5) \blue{{a}^{3}  -  {b}^{3}  = (a - b) \: ( {a}^{2}  +  {b}^{2}  + ab)}

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