Question

If p and q are roots of the quadratic equation x2-6x+5=0,then p3+q3 is

Answer 1

The value of p³ + q³ is 126.

Step-by-step explanation:

If p and q are roots of $x^2 - 6x + 5=0$,

Then,

$p+q=-\frac{\text{Coefficient of x}}{\text{Coefficient of }x^2}$

$=-\frac{-6}{1}$

= 6,

And,

$pq =\frac{\text{Constant term}}{\text{Coefficient of }x^2}}$

$=\frac{5}{1}$

= 5,

∵ (p+q)³ = p³ + q³+ 3pq(p+q)

By substituting values,

(6)³  = p³  +q³  + 3(5)(6)

216 = p³  + q³  + 90

216 - 90 = p³ + q³

$\implies p^3 + q^3 = 126$

Hence, the value of p³ + q³ is 126.

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Answer 2

Answer

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If p and q are roots of x^2 - 6x + 5= 0

By substituting values,

(6)³  = p³  +q³  + 3(5)(6)

ANSWER is attach below

216 = p³  + q³  + 90

216 - 90 = p³ + q³

Hence, the value of p³ + q³ is 126.