Question

if p/q+q/p=1; then the value of p^3+q^3​

Answer 1

Answer:

$\huge{\boxed{\pink{ {p}^{3} + {q}^{3} = 0}}}$

Step-by-step explanation:

$\bold{\underline{\red{Given}}}\longrightarrow \\ \dfrac{p}{q} + \dfrac{q}{p} = 1$

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

$\bold{\underline{\blue{Concept \: used}}}\longrightarrow \\ {x}^{3} + {y}^{3} = (x + y) \: ( {x}^{2} + {y}^{2} - xy)$

$solution \\ \: \: \: \: \: \: \: \: \dfrac{p}{q} + \dfrac{q}{p} = 1 \\ = > \dfrac{ {p}^{2} + {q}^{2} }{pq} = 1 \\ = > {p}^{2} + {q}^{2} = pq \\ = > {p}^{2} + {q}^{2} - pq = 0 \\ now \\ {p}^{3} + {q}^{3} = (p + q) \: \: ( {p}^{2} + {q}^{2} - pq \: ) \\ putting \: ( {p}^{2} + {q}^{2} - pq) = 0 \: we \: get \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (p + q) \: \: (0) \\ = > {p}^{3} + {q}^{3} = 0$

$\bold{\underline{\pink{Additional \: information}}}\longrightarrow \\ 1) {x}^{3} - {y}^{3} = (x - y) \: ( {x}^{2} + {y}^{2} + xy) \\ 2) {x}^{2} - {y}^{2} = (x + y) \: (x - y) \\ 3)(x + y) ^{2} = {x}^{2} + {y}^{2} + 2xy \\ 4) {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy$