Question

Plz Help me, Solve 15th' 2nd question and solve 16th'4th question,

Those who will give me correct answer, I will mark him Brainliest and also give 20 thanks, ​

Answer 1

Step-by-step explanation:

15 (ii)solution:-

Given

25x² + 16y² + 4z² - 40xy + 16yz - 20xz= (−5x)² + (4y)² + (2z)² + 2(−5x)(4y) + 2(4y)(2z) + 2(2z)

suitable indenties

(x+y+z)³=x³+y³+z³+2xy+2yz+2xz

Therefore,

(-5x)²+4y²+2z²+2(-5x) (2z)=(-5x+4y+2z)²

16 (iv) solution:-

1−64a³−12a+48a²

=(1)³−(4a)³−3(1)²(4a)+3(1)(4a)⅗

=(1−4a)³

=(1−4a)(1−4a)(1−4a)

Using the identity,

a³-b³-3a²b+3ab²=(a-b)³