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Question Number 49354 by peter frank last updated on 06/Dec/18

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18

$$2)x^3-qx-r=0$$$$x^3+0\times x^2+(-q)x+(-r)=0$$$$\alpha +\beta +\gamma =0$$$$\alpha \beta +\beta \gamma +\alpha \gamma =-q$$$$\alpha \beta \gamma =r$$$${(\alpha +\beta +\gamma )}^2={\alpha }^2+{\beta }^2+{\gamma }^2+2(\alpha \beta +\alpha \gamma +\beta \gamma )$$$$0=({\alpha }^2+{\beta }^2+{\gamma }^2)+2(-q)$$$$so{\alpha }^2+{\beta }^2+{\gamma }^2=2q$$$$formula{a}^3+b^3+c^3=3abcwhena+b+c=0$$$$so$$$${\alpha }^3+{\beta }^3+{\gamma }^3=3\alpha \beta \gamma =3r[since\alpha +\beta +\gamma =0]$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18

$$c)x^3=qx+r$$$${\alpha }^3=q\alpha +r$$$$mjltiplybothsideby{\alpha }^2$$$${\alpha }^5=q{\alpha }^3+r{\alpha }^2$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18

$${\alpha }^5=q{\alpha }^3+r{\alpha }^2$$$${\beta }^5=q{\beta }^3+r{\beta }^2$$$${\gamma }^5=q{\gamma }^3+r{\gamma }^2$$$$nowaddthem..$$$$6({\alpha }^5+{\beta }^5+{\gamma }^5)$$$$=6q({\alpha }^3+{\beta }^3+{\gamma }^3)+6r({\alpha }^2+{\beta }^2+{\gamma }^2)$$$$=6q\left(3r\right)+6r\left(2q\right)=30rq$$$$RHS=5({\alpha }^3+{\beta }^3+{\gamma }^3)({\alpha }^2+{\beta }^2+{\gamma }^2)$$$$=5\left(3r\right)\left(2q\right)=30rq$$$$$$

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