Question

Factorise:(p–q)3+(q–r)3+(r–p)3

Answer 1

)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

Answer 2

Answer:

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

Step-by-step explanation:

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

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

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